Powers D.L. Boundary Value Problems and Partial Differential Equations
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8 Chapter 0 Ordinary Differential Equations
Here are the details. The second derivative of u has to be replaced by its ex-
pression in terms of v, using the chain rule. Start by finding
du
dt
=
d
dz
v
z
·
dz
dt
.
Since t =1/z,alsoz =1/t,anddz/dt =−1/t
2
=−z
2
.Thus
du
dt
=−z
2
zv
−v
z
2
=−zv
+v.
Similarly we find the second derivative
d
2
u
dt
2
=
d
dz
du
dt
dz
dt
=
d
dz
(−zv
+v)
−z
2
=−z
2
(−zv
−v
+v
) =z
3
v
.
Finally, replace both terms of the differential equation:
1
z
4
z
3
v
+λ
2
v
z
=0,
or
v
+λ
2
v = 0.
This equation is easily solved, and the solution of the original is then found by
reversing the change of variables:
u(t) = t
c
1
cos(λ/t) +c
2
sin(λ/t)
. (23)
C. Second Independent Solution
Although it is not generally possible to solve a second-order linear homoge-
neous equation with variable coefficients, we can always find a second inde-
pendent solution if one solution is known. This method is called reduction of
order.
Suppose u
1
(t) is a solution of the general equation
d
2
u
dt
2
+k(t)
du
dt
+p(t)u = 0. (24)
Assume that u
2
(t) = v(t)u
1
(t) is a solution. We wish to find v(t) so that u
2
is indeed a solution. However, v(t) must not be constant, as that would not
supply an independent solution. A straightforward substitution of u
2
= vu
1
into the differential equation leads to
v
u
1
+2v
u
1
+vu
1
+k(t)(v
u
1
+vu
1
) +p(t)vu
1
=0.
Chapter 0 Ordinary Differential Equations 9
Now collect terms in the derivatives of v. The preceding equation becomes
u
1
v
+
2u
1
+k(t)u
1
v
+
u
1
+k(t)u
1
+p(t)u
1
v = 0.
However, u
1
is a solution of Eq. (24), so the coefficient of v is 0. This leaves
u
1
v
+
2u
1
+k(t)u
1
v
=0, (25)
which is a first-order linear equation for v
. Thus, a nonconstant v can be
found, at least in terms of some integrals.
Example.
Consider the equation
1 −t
2
u
−2tu
+2u = 0, −1 < t < 1,
which has u
1
(t) = t as a solution. By assuming that u
2
=v ·t and substituting,
we obtain
1 −t
2
(v
t +2v
) −2t(v
t +v) +2vt =0.
After collecting terms, we have
1 −t
2
tv
+(2 − 4t
2
)v
=0.
From here, it is fairly easy to find
v
v
=
4t
2
−2
t(1 −t
2
)
=
−2
t
+
1
1 −t
−
1
1 +t
(using partial fractions), and then
ln v
=−2ln(t) −ln(1 −t) −ln(1 +t).
Finally, each side is exponentiated to obtain
v
=
1
t
2
(1 −t
2
)
=
1
t
2
+
1/2
1 −t
+
1/2
1 +t
,
v =−
1
t
+
1
2
ln
1 +t
1 −t
.
D. Higher-Order Equations
Linear homogeneous equations of order higher than 2 — especially order 4 —
occur frequently in elasticity and fluid mechanics. A general, nth-order homo-
geneous linear equation may be written
u
(n)
+k
1
(t)u
(n−1)
+···+k
n−1
(t)u
(1)
+k
n
(t)u = 0, (26)
10 Chapter 0 Ordinary Differential Equations
Root Multiplicity Contribution
m (real) 1 ce
mt
m (real) k (c
1
+c
2
t +···+c
k
t
k−1
)e
mt
m, m (complex) 1 (a cos(βt) +b sin(βt))e
αt
m = α +iβ
m,
m (complex) k
(a
1
+a
2
t +···+a
k
t
k−1
) cos(βt)e
αt
+(b
1
+b
2
t +···+b
k
t
k−1
) sin(βt)e
αt
Tab le 3 Contributions to general solution
in which the coefficients k
1
(t), k
2
(t), etc., are given functions of t.Thetech-
niques of solution are analogous to those for second-order equations. In par-
ticular, they depend on the Principle of Superposition, which remains valid
for this equation. That principle allows us to say that the general solution
of Eq. (26) has the form of a linear combination of n independent solutions
u
1
(t), u
2
(t),...,u
n
(t) with arbitrary constant coefficients,
u(t) = c
1
u
1
(t) +c
2
u
2
(t) +···+c
n
u
n
(t).
Of course, we cannot solve the general nth-order equation (26), but we can
indeed solve any homogeneous linear equation with constant coefficients,
u
(n)
+k
1
u
(n−1)
+···+k
n−1
u
(1)
+k
n
u =0. (27)
We must now find n independent solutions of this equation. As in the second-
order case, we assume that a solution has the form u(t) = e
mt
,andfindvalues
of m for which this is true. That is, we substitute e
mt
for u in the differen-
tial equation (27) and divide out the common factor of e
mt
. The result is the
polynomial equation
m
n
+k
1
m
n−1
+···+k
n−1
m +k
n
=0, (28)
called the characteristic equation of the differential equation (27).
Each distinct root of the characteristic equation contributes as many inde-
pendent solutions as its multiplicity, which might be as high as n.Recallalso
that the polynomial equation (28) may have complex roots, which will occur
in conjugate pairs if — as we assume — the coefficients k
1
, k
2
, etc., are real.
When this happens, we prefer to have real solutions, in the form of an expo-
nential times sine or cosine, instead of complex exponentials. The contribution
of each root or pair of conjugate roots of Eq. (28) is summarized in Table 3.
Since the sum of the multiplicities of the roots of Eq. (28) is n , the sum of the
contributions produces a solution with n terms, which can be shown to be the
general solution.
Chapter 0 Ordinary Differential Equations 11
Example.
Find the general solution of this fourth-order equation
u
(4)
+3u
(2)
−4u = 0.
The characteristic equation is m
4
+3m
2
−4 =0, which is easy to solve because
it is a biquadratic. We find that m
2
=−4 or 1, and thus the roots are m =±2i,
±1, all with multiplicity 1. From Table 3 we find that a cos(2t) +b sin(2t) cor-
responds to the complex conjugate pair, m =±2i,whilee
t
and e
−t
correspond
to m = 1andm =−1. Thus we build up the general solution,
u(t) = a cos(2t) +b sin(2t) +c
1
e
t
+c
2
e
−t
.
Example.
Find the general solution of the fourth-order equation
u
(4)
−2u
(2)
+u = 0.
The characteristic equation is m
4
−2m
2
+1 =0, whose roots, found as in the
preceding, are ±1, both with multiplicity 2. From Table 3 we find that each of
the roots contributes a first-degree polynomial times an exponential. Thus, we
assemble the general solution as
u(t) = (c
1
+c
2
t)e
t
+(c
3
+c
4
t)e
−t
.
With sinh(t) = (e
t
− e
−t
)/2 and cosh(t) = (e
t
+ e
−t
)/2, the terms of the pre-
ceding combination can be rearranged to give the general solution in a differ-
ent form,
u(t) = (C
1
+C
2
t) cosh(t) +(C
3
+C
4
t) sinh(t).
Some important equations and their solutions.
1.
du
dt
=ku (k is constant),
u(t) = ce
kt
.
2.
d
2
u
dt
2
+λ
2
u =0,
u(t) = a cos(λt) +b sin(λt).
3.
d
2
u
dt
2
−λ
2
u =0,
u(t) = a cosh(λt) +b sinh(λt) or u(t) = c
1
e
λt
+c
2
e
−λt
.
4. t
2
u
+tu
−λ
2
u =0,
u(t) = c
1
t
λ
+c
2
t
−λ
.
12 Chapter 0 Ordinary Differential Equations
EXERCISES
In Exercises 1–6, find the general solution of the differential equation. Be care-
ful to identify the dependent and independent variables.
1.
d
2
φ
dx
2
+λ
2
φ = 0.
2.
d
2
φ
dx
2
−µ
2
φ = 0.
3.
d
2
u
dt
2
=0.
4.
dT
dt
=−λ
2
kT.
5.
1
r
d
dr
r
dw
dr
−
λ
2
r
2
w =0.
6. ρ
2
d
2
R
dρ
2
+2ρ
dR
dρ
−n(n + 1)R = 0.
In Exercises 7–11, find the general solution. In some cases, it is helpful to carry
out the indicated differentiation, in others it is not.
7.
d
dx
(h +kx)
dv
dx
=0(h, k are constants).
8. (e
x
φ
)
+λ
2
e
x
φ = 0.
9.
d
dx
x
3
du
dx
=0.
10. r
2
d
2
u
dr
2
+r
du
dr
+λ
2
u =0.
11.
1
r
d
dr
r
du
dr
=0.
12. Compare and contrast the form of the solutions of these three differential
equations and their behavior as t →∞.
a.
d
2
u
dt
2
+u = 0; b.
d
2
u
dt
2
=0; c.
d
2
u
dt
2
−u = 0.
In Exercises 13–15, use the “exponential guess” method to find the general
solution of the differential equations (λ is constant).
13.
d
4
u
dx
4
−λ
4
u =0.
14.
d
4
u
dx
4
+λ
4
u =0.
Chapter 0 Ordinary Differential Equations 13
15.
d
4
u
dx
4
+2λ
2
d
2
u
dx
2
+λ
4
u =0.
In Exercises 16–18, one solution of the differential equation is given. Find a
second independent solution.
16.
d
2
u
dt
2
+2a
du
dt
+a
2
u =0, u
1
(t) = e
−at
.
17. t
2
d
2
u
dt
2
+(1 − 2b)t
du
dt
+b
2
u =0, u
1
(t) = t
b
.
18.
d
dx
x
du
dx
+
4x
2
−1
4x
u =0, u
1
(x) =
cos(x)
√
x
.
In Exercises 19–21, use the indicated change of variable to solve the differential
equation.
19.
d
dρ
ρ
2
dR
dρ
+λ
2
ρ
2
R =0, R(ρ) =
u(ρ)
ρ
.
20.
d
dρ
ρ
dφ
dρ
+
4λ
2
ρ
2
−1
4ρ
φ = 0, φ(ρ) =
v(ρ)
√
ρ
.
21. t
2
d
2
u
dt
2
+kt
du
dt
+pu = 0, x = ln t, u(t) = v(x).
22. Solve each initial value problem. Assuming that the solution represents
the displacement of a mass in a mass–spring–damper system, as in the
text, describe the motion in words.
a.
d
2
u
dt
2
+4u = 0, u(0) =1,
du
dt
(0) =0;
b.
d
2
u
dt
2
+2
du
dt
+2u = 0, u(0) =1,
du
dt
(0) = 1;
c.
d
2
u
dt
2
+2
du
dt
+u = 0, u(0) =1,
du
dt
(0) = 1;
d.
d
2
u
dt
2
+2
du
dt
+0.75u = 0, u(0) =0,
du
dt
(0) =1.
23. Sheet metal is produced by repeatedly feeding the sheet between steel
rollers to reduce the thickness. In the article “On the characteristics and
mechanism of rolling instability and chatter” [Y.-J. Lin et al., J. of Manu-
facturing Scie nce and Engineering, 125 (2003): 778–786], the authors find
that the distance between rollers is well approximated by h + y,whereh
is the nominal output thickness and y is the solution of the differential
equation y
+ 2αy
+ σ
2
y = 0. The elasticity of the sheet and the rollers
provides the restoring force, and the plastic deformation of the sheet ef-
fectively provides damping.
14 Chapter 0 Ordinary Differential Equations
For high-speed operation, the system is underdamped. Solve the initial
value problem consisting of the differential equation and the initial con-
ditions y(0) =−0.001h, y
(0) = 0.
24. (Continuation) For an input speed of 25.4 m/s, it is observed that σ
∼
=
600 Hz or 1200π radians/s and α = 0.103σ . Using these values, obtain
a graph of the solution of the preceding exercise, over the range 0 < t <
0.02 s. How far does the sheet move in 0.02 s?
25. (Continuation) The damping constant α referred to in the previous ex-
ercises appears to depend on v, the speed of the sheet into the rollers,
according to the relation α/σ = A/v,whereA is a constant. From the in-
formation given previously, the value of A is about 2.62. Assuming this is
correct, find the speed v at which damping is critical.
0.2 Nonhomogeneous Linear Equations
In this section, we will review methods for solving nonhomogeneous linear
equations of first and second orders,
du
dt
=k(t)u +f (t),
d
2
u
dt
2
+k(t)
du
dt
+p(t)u = f (t).
Of course, we assume that the inhomogeneity f (t) is not identically 0. The
simplest nonhomogeneous equation is
du
dt
=f (t). (1)
This can be solved in complete generality by one integration:
u(t) =
f (t) dt +c. (2)
We have used an indefinite integral and have written c as a reminder that there
is an arbitrary additive constant in the general solution of Eq. (1). A more
precise way to write the solution is
u(t) =
t
t
0
f (z) dz +c. (3)
Here we have replaced the indefinite integral by a definite integral with vari-
able upper limit. The lower limit of integration is usually an initial time. Note
0.2 Nonhomogeneous Linear Equations 15
that the name of the integration variable is changed from t to something else
(here, z) to avoid confusing the limit with the dummy variable of integration.
The simple second-order equation
d
2
u
dt
2
=f (t) (4)
can be solved by two successive integrations.
The two theorems that follow summarize some properties of linear equa-
tions that are useful in constructing solutions.
Theorem 1. The general solution of a nonhomogeneous linear equation has the
form u(t) = u
p
(t) + u
c
(t),whereu
p
(t) is any particular solution of the nonho-
mogeneous equation and u
c
(t) is the general solution of the corresponding homo-
geneous equation.
Theorem 2. If u
p1
(t) and u
p2
(t) are p articular solutions of a differential equation
with inhomogeneities f
1
(t) and f
2
(t),respectively,thenk
1
u
p1
(t) +k
2
u
p2
is a par-
ticular solution of the differential equation w ith inhomogeneity k
1
f
1
(t) +k
2
f
2
(t)
(k
1
,k
2
are constants).
Example.
Find the solution of the differential equation
d
2
u
dt
2
+u = 1 −e
−t
.
The corresponding homogeneous equation is
d
2
u
dt
2
+u = 0,
with general solution u
c
(t) = c
1
cos(t) +c
2
sin(t) (found in Section 1). A par-
ticular solution of the equation with the inhomogeneity f
1
(t) = 1, that is, of
the equation
d
2
u
dt
2
+u = 1,
is u
p1
(t) = 1. A particular solution of the equation
d
2
u
dt
2
+u = e
−t
is u
p2
(t) =
1
2
e
−t
. (Later in this section, we will review methods for constructing
these particular solutions.) Then, by Theorem 2, a particular solution of the
given nonhomogeneous Eq. (5) is u
p
(t) = 1 −
1
2
e
−t
. Finally, by Theorem 1, the
16 Chapter 0 Ordinary Differential Equations
Inhomogeneity, f (t) Form of Trial Solution, u
p
(t)
(a
0
t
n
+a
1
t
n−1
+···+a
n
)e
αt
(A
0
t
n
+A
1
t
n−1
+···+A
n
)e
αt
(a
0
t
n
+···+a
n
)e
αt
cos(βt)
+(b
0
t
n
+···+b
n
)e
αt
sin(βt)
(A
0
t
n
+···+A
n
)e
αt
cos(βt)
+(B
0
t
n
+···+B
n
)e
αt
sin(βt)
Tab le 4 Undetermined coefficients
generalsolutionofthegivenequationis
u(t) = 1 −
1
2
e
−t
+c
1
cos(t) + c
2
sin(t).
If two initial conditions are given, then c
1
and c
2
are available to satisfy them.
Of course, an initial condition applies to the entire solution of the given dif-
ferential equation, not just to u
c
(t).
Now we turn our attention to methods for finding particular solutions of
nonhomogeneous linear differential equations.
A. Undetermined Coefficients
This method involves guessing the form of a trial solution and then finding the
appropriate coefficients. Naturally, it is limited to the cases in which we can
guess successfully: when the equation has constant coefficients and the inho-
mogeneity is simple in form. Table 4 offers a summary of admissible inhomo-
geneities and the corresponding forms for particular solution. The parameters
n,α,β and the coefficients a
0
,...,a
n
, b
0
,...,b
n
are found by inspecting the
given inhomogeneity. The table compresses several cases. For instance, f (t) in
line 1 is a polynomial if α = 0oranexponentialifn = 0andα = 0. In line 2,
both sine and cosine must be included in the trial solution even if one is absent
from f (t);butα = 0 is allowed, and so is n = 0.
Example.
Find a particular solution of
d
2
u
dt
2
+5u = te
−t
.
Weuseline1ofTable4.Evidently,n = 1andα =−1. The appropriate form
for the trial solution is
u
p
(t) = (A
0
t +A
1
)e
−t
.
When we substitute this form into the differential equation, we obtain
(A
0
t +A
1
−2A
0
)e
−t
+5(A
0
t +A
1
)e
−t
=te
−t
.
0.2 Nonhomogeneous Linear Equations 17
Now, equating coefficients of like terms gives these two equations for the coef-
ficients:
6A
0
= 1 (coefficient of te
−t
),
6A
1
−2A
0
= 0 (coefficient of e
−t
).
ThesewesolveeasilytofindA
0
=1/6, A
1
=1/18. Finally, a particular solution
is
u
p
(t) =
1
6
t +
1
18
e
−t
.
A trial solution from Table 4 will not work if it contains any term that is a so-
lution of the homogeneous differential equation. In that case, the trial solution
has to be revised by the following rule.
Revision Rule. Multiply by the lowest positive integral power of t such that no
term in the trial solut i on satisfies the corresponding homogeneous equation.
Example.
Table 4 suggests the trial solution u
p
(t) = (A
0
t + A
1
)e
−t
for the differential
equation
d
2
u
dt
2
−u = te
−t
.
However, we know that the solution of the corresponding homogeneous equa-
tion, u
−u = 0, is
u
c
(t) = c
1
e
t
+c
2
e
−t
.
The trial solution contains a term (A
1
e
−t
) that is a solution of the homoge-
neous equation. Multiplying the trial solution by t eliminates the problem.
Thus,thetrialsolutionis
u
p
(t) = t(A
0
t +A
1
)e
−t
=
A
0
t
2
+A
1
t
e
−t
.
Similarly, the trial solution for the differential equation
d
2
u
dt
2
+2
du
dt
+u = te
−t
has to be revised. The solution of the corresponding homogeneous equation
is u
c
(t) = c
1
e
−t
+c
2
te
−t
. The trial solution from the table has to be multiplied
by t
2
to eliminate solutions of the homogeneous equation.
Example: Forced Vibrations.
The displacement u(t) of a mass in a mass–spring–damper system, starting