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656 • Differential Equations
of this solves the equation, and if you have another root β, then you can add
the two solutions y = Ae
αx
and y = Be
βx
to get more solutions (try it and
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ = 0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2
π
3
−
1 = e
iπ
e
i
13
π
12
−
i = e
i
3π
2
e
i
7
π
4
i
−
i
1
θ
1 − i
2
i
−
2i
2
−
2
6
i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
see). That takes care of step 2 above.
Let’s look at step 4 next. If the two solutions to the quadratic are complex
conjugates of the form α + iβ, then by the same argument as for step 2, the
solution must be
y = Ae
(α+iβ)x
+ Be
(α−iβ)x
= e
αx
(Ae
iβx
+ Be
−iβx
),
where A and B can even be complex numbers. Now you can use Euler’s
identity (see Section 28.2 in Chapter 28) to see that
y = e
αx
(A(cos(βx) + i sin(βx)) + B(cos(βx) − i sin(βx)))
= e
αx
((A + B) cos(βx) + (A − B)i sin(βx)).
Relabel the constant (A + B) as A and the constant (A −B)i as B to get the
correct formula.
Finally, for step 3, suppose the characteristic quadratic has just one root,
α. If you substitute y = xe
αx
into the differential equation ay
00
+ by
0
+ cy = 0,
you can use y
0
= αxe
αx
+ e
αx
and y
00
= α
2
xe
αx
+ 2αe
αx
to see that
ay
00
+ by
0
+ cy = (aα
2
+ bα + c)xe
αx
+ (2aα + b)e
αx
.
If α is a double root of at
2
+ bt + c, then not only does aα
2
+ bα + c = 0, but
also 2aα + b = 0.
∗
This leads to the correct solution from step 3 above.
30.4.4 Nonhomogeneous equations and particular solutions
Now let’s see what happens if we do have some stuff in x alone, which we put
on the right-hand side. For example, consider the differential equation
y
00
− y
0
− 20y = e
x
.
This isn’t homogeneous because of the e
x
term on the right-hand side. Sup-
pose we try to guess a solution. We know that the derivatives of e
x
are all
e
x
, so let’s try y = e
x
. Then y
0
= e
x
and y
00
= e
x
, so the left-hand side
y
00
− y
0
− 20y becomes e
x
− e
x
− 20e
x
= −20e
x
. That’s not equal to the
right-hand side, but it’s pretty close. We just have to divide by −20. So, let’s
try again: set y = −
1
20
e
x
. Then y
0
and y
00
are also −
1
20
e
x
, so we have
y
00
− y
0
− 20y = −
1
20
e
x
−
−
1
20
e
x
− 20
−
1
20
e
x
= e
x
.
So we have shown that y = −
1
20
e
x
is a solution to our original equation
∗
Here’s why 2aα + b = 0 if the quadratic at
2
+ bt + c = 0 has a double root at t = α:
the discriminant is 0, so b
2
= 4ac. Then
(2aα + b)
2
= 4a
2
α
2
+ 4abα + b
2
= 4a
2
α
2
+ 4abα + 4ac = 4a(aα
2
+ bα + c) = 0.
Since (2aα + b)
2
= 0, we also have 2aα + b = 0.
Section 30.4.4: Nonhomogeneous equations and particular solutions • 657
y
00
− y
0
− 20y = e
x
. It’s not the only solution, though. To see why, consider
the related homogeneous equation
y
00
− y
0
− 20y = 0.
This was actually example (a) from Section 30.4.2 above. There we saw that
the complete solution is
y = Ae
−4x
+ Be
5x
.
So let’s play a little game. We’ll write this as y
H
instead of just y, where the
H stands for homogeneous. We have shown that
if y
H
= Ae
−4x
+ Be
5x
, then y
00
H
− y
0
H
− 20y
H
= 0.
On the other hand, we showed above that
if y
P
= −
1
20
e
x
, then y
00
P
− y
0
P
− 20y
P
= e
x
.
Here I wrote the solution −
1
20
e
x
from above as y
P
; this is called a particular
solution, which explains the subscript P . Now, suppose we add up the equa-
tions y
00
H
− y
0
H
− 20y
H
= 0 and y
00
P
− y
0
P
− 20y
P
= e
x
. Grouping derivatives
together, we get
y
00
H
+ y
00
P
− y
0
H
− y
0
P
− 20y
H
− 20y
P
= 0 + e
x
.
In fact, since the sum of the derivatives is the derivative of the sum, and the
same for the second derivative, we get
(y
H
+ y
P
)
00
− (y
H
+ y
P
)
0
− 20(y
H
+ y
P
) = e
x
.
So if y = y
H
+y
P
, then y is also a solution to our original differential equation
y
00
− y
0
− 20y = e
x
. In other words, we could take our particular solution
y
P
= −
1
20
e
x
,
which actually solves the differential equation; then add any solution to the
homogeneous version of the differential equation; the result is still a solution
to the original differential equation. Furthermore, all the solutions to the
nonhomogeneous equation are in this form.
The same methodology works for both the first-order and the second-
order cases. The only issue is how to guess the particular solution. In the
next section, we’ll see how to make a guess of what the form of the solution
should be (this is similar to the partial fraction technique from Section 18.3
of Chapter 18). Then if you’re lucky, you can plug in that form and find the
unknown constants in order to nail down the particular solution.
658 • Differential Equations
Here’s a summary of our methods so far:
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f (
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f (
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f (
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f (
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r =
1 + cos(θ)
r =
1 +
3
4
cos(
θ)
−
1
4
r =
sin(2θ)
r =
sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1
+ sin(θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
appro
ximating region
exact
region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ =
0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1
= e
0
e
i
π
4
i = e
i
π
2
e
i
2
π
3
−
1 = e
iπ
e
i
13
π
12
−
i = e
i
3π
2
e
i
7
π
4
i
−
i
1
θ
1 − i
2
i
−
2i
2
−
2
6
i
−
6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f (x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
1. Rearrange the equation into the correct form. That is, put all the x-junk
on the right-hand side. You should be able to reduce the equation to
dy
dx
+ ay = f(x)
for the first-order case, or
a
d
2
y
dx
2
+ b
dy
dx
+ cy = f (x)
for the second-order case.
2. Using the techniques from Sections 30.4.1 and 30.4.2 above, solve the
associated homogeneous equation
dy
dx
+ ay = 0 or a
d
2
y
dx
2
+ b
dy
dx
+ cy = 0.
The solution, which we’ll write as y
H
, will have one or two undetermined
constants in it (depending on whether the equation is first- or second-
order). We call y
H
the homogeneous solution of the equation.
3. If the original function f is actually 0, then we’re already done; the
complete solution is y = y
H
.
4. On the other hand, if the function f is anything other than 0, then
write down the form for the particular solution y
P
(see Section 30.4.5
below). The form will have some constants which must be determined.
Substitute y
P
into the original equation and equate coefficients to find
the constants.
5. Finally, the solution is y = y
H
+ y
P
.
We’ll look at what happens if you are dealing with an initial value problem
(IVP) in Section 30.4.8 below. Meanwhile, let’s see how to find a particular
solution.
30.4.5 Finding a particular solution
So far, we have blissfully ignored the stuff involving x which could appear on
the right-hand side (it was called f(x) earlier). Now it’s time to deal with it.
The tactic is to write down the form of the particular solution, then to find
the actual solution by plugging the form into the equation. The table on the
next page shows how to come up with the correct form. For example, in the
differential equation
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r =
1 + cos(θ)
r =
1 +
3
4
cos(
θ)
−
1
4
r =
sin(2θ)
r =
sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1
+ sin(θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
y
0
− 3y = 5e
2x
,
the right-hand side is a multiple of e
2x
; so the table indicates that the form
should be y
p
= Ce
2x
, where C is a constant that we have to find by substi-
tuting y
P
into the original equation. It’s easy to see that y
0
P
= 2Ce
2x
, so we
have
2Ce
2x
− 3(Ce
2x
) = 5e
2x
.
This reduces to −Ce
2x
= 5e
2x
, so C = −5. The particular solution is therefore
y
P
= −5e
2x
. In fact, since we saw in Section 30.4.1 above that the solution
Section 30.4.5: Finding a particular solution • 659
to the homogeneous version y
0
−3y = 0 is y
H
= Ae
3x
, we now know that the
full solution to y
0
− 3y = 5e
2x
is
y = y
H
+ y
P
= Ae
3x
− 5e
2x
,
where A is an unknown constant. Note that the homogeneous solution involves
an unknown constant, while the particular solution must have no unknown
constants.
Now, here’s the table:
If f is a . . . then the form is . . .
polynomial of degree n y
P
= general polynomial
of degree n
e.g., f(x) = 7 y
P
= a
f(x) = 3x − 2 y
P
= ax + b
f(x) = 10x
2
y
P
= ax
2
+ bx + c
f(x) = −x
3
− x
2
+ x + 22 y
P
= ax
3
+ bx
2
+ cx + d
multiple of an exponential e
kx
y
P
= Ce
kx
e.g., f(x) = 10e
−4x
y
P
= Ce
−4x
f(x) = e
x
y
P
= Ce
x
multiple of cos(kx) y
P
= C cos(kx) + D sin(kx)
+ multiple of sin(kx)
e.g., f(x) = 2 sin(3x) − 5 cos(3x) y
P
= C cos(3x) + D sin(3x)
f(x) = cos(x) y
P
= C cos(x) + D sin(x)
f(x) = 2 sin(11x) y
P
= C cos(11x) + D sin(11x)
a sum or product of one of the sum or product of forms
the above (if a product, omit a constant)
e.g., f(x) = 2x
2
+ e
−6x
y
P
= ax
2
+ bx + c + Ce
−6x
f(x) = 2x
2
e
−6x
y
P
= (ax
2
+ bx + c)e
−6x
f(x) = 7e
2x
sin(3x) y
P
= (C cos(3x) + D sin(3x))e
2x
f(x) = cos(2x) + 6 sin(x) y
P
= C cos(2x) + D sin(2x)
+ E cos(x) + F sin(x)
f(x) = 4x cos(3x) y
P
= (x + b)(C cos(3x) + D sin(3x))
If y
P
conflicts with y
H
, multiply the form by x or x
2
as appropriate.
This table should be fairly self-explanatory, except for the last line, which
will be explained in Section 30.4.7 below, and also the instruction “if a prod-
uct, omit a constant.” To see what this instruction means, first note that
there is a redundant constant if you just multiply two forms together. For ex-
ample, 2x
2
e
−6x
looks as if it should lead to the form (ax
2
+ bx + c)Ce
−6x
, but
the constant C is unnecessary and can be omitted, since it can be absorbed
into the other constants a, b, and c. The same sort of thing applies to the
examples 7e
2x
sin(3x) and 4x cos(3x) in the above table.
(By the way, the table only shows you what to do if f happens to be a
polynomial, an exponential, a sine, a cosine, or some product or sum of one
or more of these types of function. Otherwise the method just doesn’t work.
There is a fancier method called “variation of parameters” which is much
more general, but it’s outside the scope of this book.)
660 • Differential Equations
30.4.6 Examples of finding particular solutions
Once you’ve written down the form for y
P
, you still have to substitute y
P
into the original differential equation in order to find the constants. To make
the calculation easier, you should first calculate y
0
P
and y
00
P
(for the first order
case, you actually only need y
0
P
). Let’s look at one example of this; then we’ll
finally go back and complete the two unresolved examples from Section 30.4
above.
First consider the differential equation
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r =
1 + cos(θ)
r =
1 +
3
4
cos(
θ)
−
1
4
r =
sin(2θ)
r =
sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1
+ sin(θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
appro
ximating region
exact
region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ =
0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1
= e
0
e
i
π
4
i = e
i
π
2
e
i
2
π
3
−
1 = e
iπ
e
i
13
π
12
−
i = e
i
3π
2
e
i
7
π
4
i
−
i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
y
00
− 4y
0
+ 4y = 25e
3x
sin(2x).
Let’s quickly dispense with the homogeneous part; in fact, the characteristic
quadratic equation for y
00
− 4y
0
+ 4y = 0 is t
2
− 4t + 4 = 0, which has one
solution, namely t = 2. So we have y
H
= Ae
2x
+ Bxe
2x
, where A and B are
constants. Now let’s look for a particular solution. Break up the right-hand
side of our differential equation, 25e
3x
sin(2x), into two components: 25e
3x
and sin(2x). According to the above table, the form for a constant multiple
of e
3x
is Ce
3x
; and the form for sin(2x) is C cos(2x) + D sin(2x). We need to
multiply these together, but we can consolidate the constants as we do so and
write
y
P
= e
3x
(C cos(2x) + D sin(2x))
as our form. Now, let’s do some fiddly calculations using the product rule
many times:
y
P
= e
3x
(C cos(2x) + D sin(2x)),
y
0
P
= e
3x
(−2C sin(2x) + 2D cos(2x)) + 3e
3x
(C cos(2x) + D sin(2x))
= e
3x
((3C + 2D) cos(2x) + (3D − 2C) sin(2x)),
y
00
P
= e
3x
(−2(3C + 2D) sin(2x) + 2(3D − 2C) cos(2x))
+ 3e
3x
((3C + 2D) cos(2x) + (3D − 2C) sin(2x))
= e
3x
((5C + 12D) cos(2x) + (5D − 12C) sin(2x)).
Now it’s time to substitute this mess into the original differential equation
y
00
− 4y
0
+ 4y = 25e
3x
sin(2x). We get the gross-looking equation
e
3x
((5C + 12D) cos(2x) + (5D − 12C) sin(2x))
− 4e
3x
((3C + 2D) cos(2x) + (3D − 2C) sin(2x))
+ 4e
3x
(C cos(2x) + D sin(2x)) = 25e
3x
sin(2x),
which mercifully simplifies to
e
3x
(4D − 3C) cos(2x) + e
3x
(−4C − 3D) sin(2x) = 25e
3x
sin(2x).
To make these expressions equal for all x, the e
3x
cos(2x) stuff has to disappear
and the coefficient of e
3x
sin(2x) must be 25. This means that 4D − 3C = 0
and −4C − 3D = 25. Solving these equations simultaneously, you should get
C = −4 and D = −3. We now know that y
P
= e
3x
(−4 cos(2x) − 3 sin(2x)),
so the complete solution is
y = y
H
+ y
P
= Ae
2x
+ Bxe
2x
− e
3x
(4 cos(2x) + 3 sin(2x)),
Section 30.4.6: Examples of finding particular solutions • 661
where A and B are constants.
Now it’s time to finish off two of the examples from Section 30.4 above, as
promised:
y
0
+ 2y = 4x +
1
3
sin(5x) and y
00
− 5y
0
+ 6y = 2x
2
e
x
.
At this point, you should try to solve them both. Once you’ve done that, read
on.
The left-hand example is a first-order equation. The homogeneous version
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ = 0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2
π
3
−
1 = e
iπ
e
i
13
π
12
−
i = e
i
3π
2
e
i
7
π
4
i
−
i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
is y
0
+2y = 0, which has the solution y = Ae
−2x
, where A is a constant. Upon
consulting the above table, we see that the form for a particular solution is
y
P
= ax + b + C cos(5x) + D sin(5x). We’ll need the derivative, namely
y
0
P
= a − 5C sin(5x) + 5D cos(5x). Substituting y
0
P
and y
P
into the original
equation, we get
(a−5C sin(5x)+5D cos(5x))+2(ax+b+C cos(5x)+D sin(5x)) = 4x+
1
3
sin(5x),
which reduces to
2ax + 2b + a + (5D + 2C) cos(5x) + (2D − 5C) sin(5x) = 4x +
1
3
sin(5x).
Now we have to equate coefficients of various components of this expression.
The coefficient of x is 2a on the left-hand side and 4 on the right-hand side,
so a = 2. The constant coefficient on the left is 2b + a, whereas there’s no
constant on the right, so 2b + a = 0. This means that b = −1. Meanwhile,
there’s no term in cos(5x) on the right, so 5D + 2C = 0. On the other hand,
the sin(5x) terms must match, so we have 2D − 5C = 1/3. Solving these last
two equations simultaneously (try it!) gives C = −5/87 and D = 2/87. So,
we have
y
P
= 2x − 1 −
5
87
cos(5x) +
2
87
sin(5x);
putting it all together, we get the solution
y = y
H
+ y
P
= Ae
−2x
+ 2x − 1 −
5
87
cos(5x) +
2
87
sin(5x),
where A is a constant.
How about the other example above? That’s a second-order equation,
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
with homogeneous version given by y
00
− 5y
0
+ 6y = 0. The characteristic
quadratic equation is t
2
−5t + 6 = 0, which has solutions t = 2 and t = 3. So,
y
H
= Ae
2x
+ Be
3x
, where A and B are constants. Now it’s time to deal with
the particular solution. Since the right-hand side of the original differential
equation is 2x
2
e
x
, the form should be y
P
= (ax
2
+ bx + c)e
x
; remember that
you don’t need a constant outside of the e
x
, since that constant could be
absorbed into a, b and c. Let’s differentiate y
P
a couple of times:
y
P
= (ax
2
+ bx + c)e
x
,
y
0
P
= (ax
2
+ bx + c)e
x
+ (2ax + b)e
x
= (ax
2
+ (2a + b)x + (b + c))e
x
,
y
00
P
= (ax
2
+ (2a + b)x + (b + c))e
x
+ (2ax + (2a + b))e
x
= (ax
2
+ (4a + b)x + (2a + 2b + c))e
x
.
662 • Differential Equations
Now substitute into the original equation y
0
− 5y
0
+ 6y = 2x
2
e
x
to get
(ax
2
+ (4a + b)x + (2a + 2b + c))e
x
− 5(ax
2
+ (2a + b)x + (b + c))e
x
+ 6(ax
2
+ bx + c)e
x
= 2x
2
e
x
.
This simplifies to
(2ax
2
+ (−6a + 2b)x + (2a − 3b + 2c))e
x
= 2x
2
e
x
.
Now equate coefficients to see that 2a = 2, −6a+ 2b = 0 and 2a−3b + 2c = 0.
This means that a = 1, b = 3 and c =
7
2
, so y
P
= (x
2
+ 3x +
7
2
)e
x
. The
solution to the whole equation is therefore
y = y
H
+ y
P
= Ae
2x
+ Be
3x
+
x
2
+ 3x +
7
2
e
x
,
where A and B are constants.
30.4.7 Resolving conflicts between y
P
and y
H
The last line of the table in Section 30.4.5 above indicates that there might
be conflicts between y
P
and y
H
. How can this happen? Well, consider the
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ = 0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
differential equation
y
00
− 3y
0
+ 2y = 7e
2x
.
The homogeneous version is y
00
− 3y
0
+ 2y = 0, with characteristic quadratic
equation given by t
2
−3t + 2 = (t −1)(t−2) = 0, so the homogeneous solution
is
y
H
= Ae
x
+ Be
2x
.
Here A and B are unknown constants. Now, since the right-hand side of the
differential equation is 7e
2x
, our table says that the form for the particular
solution is y
P
= Ce
2x
. The sad fact, alas, is that this choice will crash and
burn. Indeed, this y
P
is included in y
H
by setting A = 0 and B = C. This
means that if you plug y
P
= Ce
2x
into the differential equation, you will get 0
on the left-hand side (try it!), so it doesn’t work. Instead, as the final line of
the table indicates, you need to introduce an extra power of x to make it work.
So, we’ll use y
P
= Cxe
2x
instead. Let’s see what happens now. First, note
that y
0
P
= 2Cxe
2x
+ Ce
2x
and y
00
P
= 4Cxe
2x
+ 4Ce
2x
, so when you substitute
into the differential equation above, you get
(4Cxe
2x
+ 4Ce
2x
) − 3(2Cxe
2x
+ Ce
2x
) + 2Cxe
2x
= 7e
2x
.
The terms in xe
2x
cancel completely, and you’re left with Ce
2x
= 7e
2x
. So
C = 7, meaning that y
P
= 7xe
2x
. Finally, the complete solution is given by
y = y
H
+ y
P
= Ae
x
+ Be
2x
+ 7xe
2x
.
One more example. If you want to solve
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−4
−5
4
5
f(θ)
f(θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
y
00
+ 6y
0
+ 9y = e
−3x
,
you’ll have to go even further than before. Now the homogeneous equation
y
00
+ 6y
0
+ 9y = 0 has characteristic quadratic t
2
+ 6t + 9 = (t + 3)
2
, so
Section 30.4.8: Initial value problems (constant-coefficient linear) • 663
the homogeneous solution is y
H
= Ae
−3x
+ Bxe
−3x
. Since the right-hand
side of the differential equation is e
−3x
, we’d want to take y
P
= Ce
−3x
.
That won’t work, since it’s included in y
H
(with A = C and B = 0). Even
y
P
= Cxe
−3x
won’t work, since that’s also included in y
H
(with A = 0 and
B = C). So we have to go all the way up to x
2
and set y
P
= Cx
2
e
−3x
.
Now you can differentiate twice to see that y
0
P
= 2Cx
−3x
− 3Cx
2
e
−3x
and
y
00
P
= 2Ce
−3x
−12Cxe
−3x
+ 9Cx
2
e
−3x
(check this!). I leave it to you to plug
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ = 0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2
π
3
−
1 = e
iπ
e
i
13
π
12
−
i = e
i
3π
2
e
i
7
π
4
i
−
i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
these quantities into the original equation and show that it all simplifies to
2Ce
−3x
= e
−3x
. This means that C =
1
2
, so the solution to the differential
equation is y = y
H
+ y
P
= Ae
−3x
+ Bxe
−3x
+
1
2
x
2
e
−3x
for some constants A
and B.
30.4.8 Initial value problems (constant-coefficient linear)
Let’s see how to deal with initial-value problems (IVPs) involving constant-
coefficient linear differential equations. As usual, to solve an IVP, first solve
the differential equation, then use the initial conditions to find the remaining
unknown constants.
Let’s modify the last two examples from Section 30.4.6 above to make
them into IVPs, then solve them. For the first example, suppose you are
given that y
0
+ 2y = 4x +
1
3
sin(5x), and that y(0) = −1. Well, ignoring the
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ = 0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
condition y(0) = −1 for the moment, we already saw that the general solution
is
y = Ae
−2x
+ 2x − 1 −
5
87
cos(5x) +
2
87
sin(5x).
Now we also know that y(0) = −1, which means that when x = 0, y = −1.
Substituting this in, we get
−1 = Ae
0
+ 2(0) − 1 −
5
87
cos(0) +
2
87
sin(0) = A − 1 −
5
87
.
This reduces to A = 5/87, so the solution to the IVP is
y =
5
87
e
−2x
+ 2x − 1 −
5
87
cos(5x) +
2
87
sin(5x).
There are no unknown constants.
To modify the second example, let’s suppose that y
00
− 5y
0
+ 6y = 2x
2
e
x
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2π
r = |1 + 2 cos(θ)|
2i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
and that y(0) = y
0
(0) = 0. As we saw in Section 30.4.6, the general solution
(ignoring the initial conditions y(0) = 0 and y
0
(0) = 0) is given by
y = Ae
2x
+ Be
3x
+
x
2
+ 3x +
7
2
e
x
.
We’ll need to differentiate this once to take advantage of the fact that we
know what y
0
(0) is; check that
y
0
= 2Ae
2x
+ 3Be
3x
+
x
2
+ 5x +
13
2
e
x
.
So, when x = 0, we know that both y and y
0
are equal to 0; substituting into
the equation for y gives
0 = Ae
0
+ Be
0
+
0
2
+ 3(0) +
7
2
e
0
= A + B +
7
2
,
664 • Differential Equations
whereas substituting into the equation for y
0
gives
0 = 2Ae
0
+ 3Be
0
+
0
2
+ 5(0) +
13
2
e
0
= 2A + 3B +
13
2
.
Solving these equations simultaneously, we get A = −4 and B =
1
2
. This
means that the solution to the IVP is
y = −4e
2x
+
1
2
e
3x
+
x
2
+ 3x +
7
2
e
x
.
Notice that in both examples there are no constants left: the initial conditions
have allowed us to home in on the unique solution. Without initial conditions,
there will always be one or two unknown constants.
Let’s look at one last IVP example. Suppose that
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ = 0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2
π
3
−
1 = e
iπ
e
i
13
π
12
−
i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
y
00
+ 6y
0
+ 13y = 26x
3
− 3x
2
− 24x, y(0) = 1, y
0
(0) = 2.
The homogeneous equation is y
00
+6y
0
+13y = 0, with characteristic quadratic
equation t
2
+ 6t + 13 = 0. Using the quadratic formula, the solutions to this
last equation are t = (−6 ±
√
36 − 4 · 13)/2 = −3 ± 2i. This means that
y
H
= e
−3x
(A cos(2x) + B sin(2x)). Turning now to the particular solution:
since the right-hand side (the x-stuff) of the original equation is a cubic, we
should write down the form y
P
= ax
3
+ bx
2
+ cx + d. Now we have to find the
constants a through d by substituting y
P
into the differential equation. Note
that y
0
P
= 3ax
2
+ 2bx + c and y
00
P
= 6ax + 2b. Substituting, we get
(6ax + 2b) + 6(3ax
2
+ 2bx + c) + 13(ax
3
+ bx
2
+ cx + d) = 26x
3
−3x
2
−24x.
Equating coefficients (just as we did for partial fractions) for x
3
, x
2
, x, and 1,
we get 13a = 26, 18a +13b = −3, 6a +12b+ 13c = −24, and 2b +6c+ 13d = 0,
respectively. I leave it to you to solve these equations and see that a = 2,
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hyp
otenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
even smaller
y = g(
x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t = 0
t = π/
6
t = π/
4
t = π/
3
t = π/
2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t = 0
t = 1
/2
t = 3
/2
t = 2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r = 3 sin(
θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r = 1 + 2 cos(
θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r = 1 + cos(
θ)
r = 1 +
3
4
cos(
θ)
−
1
4
r = sin(2
θ)
r = sin(3
θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1 + sin(
θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
approximating region
exact region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3i
−1
θ = 0
θ =
π
4
θ =
π
2
θ =
2π
3
θ = π
θ =
13π
12
θ =
3π
2
θ =
7π
4
1 = e
0
e
i
π
4
i = e
i
π
2
e
i
2π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
b = −3, c = 0, and d = 6/13. So y
P
= 2x
3
− 3x
2
+ 6/13, and therefore
y = y
H
+ y
P
= e
−3x
(A cos(2x) + B sin(2x)) + 2x
3
− 3x
2
+
6
13
for some constants A and B. Now, to find these constants, let’s use the initial
conditions. Since y(0) = 1, we know that y = 1 when x = 0; substituting, we
have
1 = e
−3(0)
(A cos(0) + B sin(0)) + 2(0)
3
− 3(0)
2
+
6
13
= A +
6
13
,
so A = 7/13. Meanwhile, differentiating the expression for y gives
y
0
= e
−3x
(−2A sin(2x)+2B cos(2x))−3e
−3x
(A cos(2x)+B sin(2x))+6x
2
−6x.
Now, since y
0
(0) = 2, we know that y
0
= 2 when x = 0; substituting this into
the above expression for y
0
, we get
2 = e
0
(−2A sin(0) + 2B cos(0)) − 3e
0
(A cos(0) + B sin(0)) + 6(0)
2
− 6(0)
= 2B − 3A.
Section 30.5: Modeling Using Differential Equations • 665
Since A = 7/13, we can solve this last equation to find that B = 47/26. Now
we plug these values in to find the final answer:
y = e
−3x
7
13
cos(2x) +
47
26
sin(2x)
+ 2x
3
− 3x
2
+
6
13
.
Once again, note that there are no constants involved here: the initial condi-
tions (that is, the values of y(0) and y
0
(0)) pinpoint the explicit solution.
30.5 Modeling Using Differential Equations
Many quantities in the real world can be modeled (that is, theoretically
approximated) by differential equations. Examples include heat flow, wave
height, inflation, current in electrical circuits, and population growth, to name
a few. Here’s a simple example of a somewhat realistic situation involving
population growth.
A certain culture of bacteria grows exponentially in such a way that its
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hyp
otenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(
x)
a
b
a + ε
ε
Z
b
a
+ε
f(x) dx
small
ev
en smaller
y = g(
x)
infinite
area
finite
area
1
y =
1
x
y =
1
x
p
, p
< 1 (typical)
y =
1
x
p
, p
> 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(
x)
(
a, f(a))
a
−
1
0
1
a
6
1
2
7
1
2
7
?
−
2
−
1
−
2
t =
0
t = π
/6
t = π
/4
t = π
/3
t = π
/2
3
0
t = −
2
t = −
3/2
t = ±
1
t = −
1/2
t =
0
t =
1/2
t =
3/2
t =
2
12
−
12
θ
r
P
θ
r
P
11
π
6
2
(
−1, −1)
wrong
point
π
4
5
π
4
√
2
(0
, 1)
(0
, −3)
(
−2, 0)
π
2
3
π
2
π
r =
3 sin(θ)
3
π
2
θ
2
π
1
0
−
1
−
2
−
3
0
3
2
−
3
2
0
r =
1 + 2 cos(θ)
2
π
3
4
π
3
0
π
0
pi
−
3
2
3
π
2
1
2
3
0
−
1
−
2
−
3
0 ≤ θ ≤
2
π
3
0 ≤ θ ≤ π
0 ≤ θ ≤ 2
π
r =
1 + cos(θ)
r =
1 +
3
4
cos(
θ)
−
1
4
r =
sin(2θ)
r =
sin(3θ)
r =
1
π
θ
0 ≤ θ ≤ 4
π
r =
2
1
+ sin(θ)
−
π
4
≤ θ ≤
5
π
4
0 ≤ θ ≤ 2
π
0 ≤ θ ≤ π
−
4
−
5
4
5
f(
θ)
f(
θ + dθ)
θ
dθ
θ + dθ
appro
ximating region
exact
region
0 ≤ θ ≤ 2
π
r = |
1 + 2 cos(θ)|
2
i
2 − 3
i
−
1
θ =
0
θ =
π
4
θ =
π
2
θ =
2
π
3
θ = π
θ =
13
π
12
θ =
3
π
2
θ =
7
π
4
1
= e
0
e
i
π
4
i = e
i
π
2
e
i
2
π
3
−1 = e
iπ
e
i
13π
12
−i = e
i
3π
2
e
i
7π
4
i
−i
1
θ
1 − i
2i
−2i
2
−2
6i
−6i
6
−6
−
√
3
R
ϕ
2
1/5
θ =
π
6
θ =
17π
30
θ =
29π
30
θ =
41π
30
θ =
53π
30
z
0
z
1
z
2
z
3
z
4
−
√
3
2
√
3
2
1
2
i
−i
19π
6
−i
7π
6
i
5π
6
i
17π
6
i
29π
6
ln(2)
−
7π
4
−
3π
4
π
4
5π
4
9π
4
3
2
i
0
1
2
3
4
dx
y
x
y =
p
1 − (x − 3)
2
2πx
a
b
y = f(x)
A
B
y =
√
x
1
y = 2x
3
y = x
4
(2, 16)
−5
5
6
y = h
y−h
h−y
x = h
y
x−h
radius of shell = x−h
h−x
radius of shell = h−x
8
P
h
P
(slice)
(axis)
l
L
1
2
Base
Cross-section
Area = A
Area = A(x)
y = e
x
A
B
dx
dy
x + dx
a
b
p
(dx)
2
+ (dy)
2
instantaneous hourly rate of increase is equal to twice the number of bacteria
in the culture. Suppose that an antibiotic is continuously introduced into the
culture at the constant rate of 8 ounces per hour. Each ounce of antibiotic
present kills 25,000 of the bacteria per hour. What is the minimum initial
population of bacteria that need to be present in order to ensure that the
culture is never completely wiped out?
The idea here is that the number of bacteria is increasing as they breed,
but the amount of killer antibiotic is increasing too as it gets pumped into the
petri dish. Which one wins, the bacteria or the antibiotic? To find out, we
need to write down a differential equation that models the situation. In effect,
we have to translate the word problem into a differential equation. If there
were no antibiotic, you’d have the standard population growth differential
equation with k = 2:
dP
dt
= 2P,
where P is the population at time t hours. (We looked at this sort of thing in
Section 9.6.1 of Chapter 9.) Now we have to modify this to take the antibiotic
into account. At time t hours, we know there are 8t ounces of antibiotic
present, so the death rate due to this amount present is 8t ×25000 = 200000t.
The correct differential equation is therefore
dP
dt
= 2P − 200000t.
This can be rearranged into standard form as
dP
dt
− 2P = −200000t.
The integrating factor (see Section 30.3 above) for this first-order linear equa-
tion is e
R
−2 dt
, which simplifies to e
−2t
. Multiplying the equation by the
integrating factor, we get
e
−2t
dP
dt
− 2e
−2t
P = −200000e
−2t
t.