Дыхта В.А. Динамические системы в экономике. Введение в анализ одномерных моделей
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y
0
= y, y(0) = 1
y
0
+ 3y = 12, y(0) = 10
2y
0
+
1
2
y = 12, y(0) = 10
y
0
= 5, y(0) = 1
y
0
= 6y − 6, y(0) = 3
y
0
+ 2y = 4, y(0) = 3
y(1) = 1
y
0
= 5 − y; y
0
= t − y; y
0
= y − t
2
ty
0
+ (1 − t)y = e
2t
y
0
= y
3
/t
3
; y
0
= t
3
/y
3
y
0
= −
1
t
y + sin t, t > 0
t
2
y
0
+ 3ty = (sin t)/t, t < 0
ty
0
+ 2y = e
t
, t > 0
y
0
+ (tg t)y = t sin 2t, −
π
2
< t <
π
2
ty
0
+ 2y = t
2
− t + 1, y(1) =
1
2
ty
0
+ y = e
t
, y(1) = 1
ty
0
+ 2y = sin t, y(π) =
1
π
y
0
+ (ctg t)y = 4 sin t, y(−
π
2
) = 0
t(2 + t)y
0
+ 2(1 + t)y = 1 + 3t
2
, y(−1) = 1
y
0
+ y = 1/(1 + t
2
), y(0) = 0
t = 0
t > 0 t → 0
y
0
+
2
t
y =
1
t
2
; y
0
−
1
t
y = t
y
0
+
1
t
y = t
1/2
; y
0
+
1
t
y = (cos t)/t
y
0
− 2ty = 1, y(0) = y
0
y = e
t
2
·
√
π
2
(t) + y
0
¸
t
2
y
0
+ 2ty − y
3
= 0
y
0
= ry − ky
2
y
0
= ay − by
3
, a, b > 0
y
0
= t
2
/2 y
0
= t
2
/y(1 + t
3
)
y
0
+ y
2
sin t = 0 y
0
= (1 + t + y
2
+ ty
2
)
y
0
= (cos
2
t)(cos
2
2y) ty
0
= (1 − y
2
)
1/2
dy
dt
=
t−e
−t
y +e
y
dy
dt
=
t
2
1+y
2
(sin 2t) dt + (cos 3y) dy = 0, y(π/2) = π/3
t dt + ye
−t
dy = 0, y(0) = 1
dx
dt
=
r
2
t
, x(1) = 2
y
0
= 2t/(y + t
2
y), y(0) = −2
y
0
= ty
3
(1 + t
2
)
−1/2
, y(0) = 1
y
0
= 2t/(1 + 2y), y(2) = 0
y
0
= t(t
2
+ 1)/4y
3
, y(0) = −1/
√
2
y
0
= 3t
2
/(3y
2
− 4), y(1) = 0
dt/dy = 0
y
2
(1 − t
2
)
1/2
dy = (sin t)
−1
dt
−1 < t < 1
dy
dt
=
at + b
ct + d
,
a b c d
dy
dt
=
ay + b
cy + d
,
a b c d
dy
dt
=
y − 4t
t − y
,
x = y/t
x t
y
0
= f
³
y
t
´
,
y/t
f(x)
(a, b)
(t, y)
y = at y = bt
x = y/t
x = ¯x = ¯x f(x) = x
y
0
=
t+y
t
y
0
= 2y dt −t dy
y
0
=
t
2
+ty +y
2
t
2
y
0
=
t
2
+3y
2
2ty
y
0
=
4y−3 t
2t−y
y
0
= −
4t+3y
2t+y
y
0
=
t+3y
t−y
(t
2
+ 3ty + y
2
) dt − t
2
dy = 0
y
0
=
2y − t
2t − y
.
y
0
=
2y − t + 5
2t − y − 4
.
(t, y) (τ, x) t = τ − a x =
y − b a b
1
y
0
= −
4t+3y +15
2t+y +7
2
y
0
=
t+3y − 5
t−y− 1
y
0
= f(t, y)
f
f(t, αt) = f (1, t),
α
y
0
=
t
3
+ty + y
3
t
2
y+ty
2
; y
0
= ln t − ln y +
t+y
t−y
y
0
=
(t
2
+3ty +4 y
2
)
1/2
t+2y
; y =
sin ty
t
2
+y
2
y
0
= f(t, y)
f(t, y) = −
P (t, y)
Q(t, y)
,
P (t, y) dt + Q(t, y) dy = 0.
F (t, y)
F
0
t
≡ P (t, y) F
0
y
≡ Q(t, y)
ω = P dt + Q dy
dF ≡ ω
P Q D
(t, y)
D
P
0
y
= Q
0
t
D.
P Q
y(t)
F (t, y(t)) = C
C F
F (t, y) = C,
C
F
0
t
(t, y) = P (t, y)
F (t, y) =
Z
P (t, y) dt + ϕ(y),
ϕ(y)
t y
ϕ(y) F
0
y
(t, y) =
Q(t, y)
ϕ
0
(y) = Q −
∂
∂y
Z
P dt.
t
ϕ
y
ϕ F
F (t, y) =
Z
P (t, y) dt +
Z
·
Q(t, y) −
Z
P (t, y) dt
¸
dy. 2
(2t + 3) + (2y − 2)y
0
= 0
(3t
2
− 2ty + 2) dt + (6y
2
− t
2
+ 3) dy = 0
(2ty
2
+ 2y) + (2t
2
y + 2t)y
0
= 0
(2t + 4y) + (2t − 2y)y
0
= 0
dy
dt
= −
at+by
bt+cy
dy
dt
= −
at−by
bt−cy
(e
t
sin y − 2y sin t) dt + (e
t
cos y + 2 cos t) dy = 0
(e
t
sin y + 3y) dt − (3t − e
t
sin y) dy = 0
(ye
ty
cos 2t − 2te
ty
sin 2t + 2t) dt + (te
ty
cos 2t − 3) dy = 0
(y/t + 6t) dt + (ln t − 2) dy = 0, t > 0
t ln y + ty) dt + (y ln t + ty) dy = 0, t, y > 0
t dt
(t
2
+y
2
)
3/2
+
y dy
(t
2
+y
2
)
3/2
= 0
D
D
D
g
t
(t, y) ≡ D g
g t D D
D
x
t
+ [ln(tx) + 1] x
0
= 0, t, x > 0
(2t − y) dt + (2y − t) dy = 0, y(1) = 3;
(3t
2
+ y + 1) dt − (4y − t) dy = 0, y(1) = 0;
a
a
(ty
2
+ at
2
y) dt + (t + y)t
2
dy = 0;
(ye
2ty
+ t) dt + ate
2ty
dy = 0.
P dt+Q dy = 0
m(t, y) 6= 0
m(t, y) × [P (t, y) dt + Q(t, y) dy] = 0
m(t, y) 6= 0
m(t, y) = ( ty
2
)
−1
(y
2
+ ty) dt − t
2
dy = 0,
P m
0
y
− Qm
0
t
+ (P
y
− Q
t
)m = 0.
m(t, y)
P dy + Q dy = 0
t
y
t
m(t)
dm
dt
=
P
0
y
− Q
0
t
Q
m.
t m
t
m(t)
(3ty + y
2
) + (t
2
+ ty)
dy
dt
= 0
m(t)
y
dy
dt
= ay + b(t)
e
−at
y
0
= −2y + e
−t
, y(0) =
3
4
.
y
0
= a(t)y + b(t)
m(t) = exp(−
Z
a(t) dt).
1
) y
0
= 2y + t
2
e
2t
;
2
y
0
= −
1
t
y + 3 cos 2t, t > 0.
t
2
y
3
+ t(1 + y
2
)y
0
= 0, m(t, y) = 1/ty
3
³
sin y
y
− 2e
−t
sin t
´
dt +
³
cos y+2e
−t
cos t
t
´
dy = 0, m(t, y) = e
t
y
y dy + (2t − ye
y
) dy = 0, m(t, y) = y
(t + 2) sin y dt + t cos y dy = 0, m(t, y) = te
t
M = (Q
0
t
− P
0
y
)/Q
y P + Qy
0
m(y) = exp
µ
Z
M(y) dy
¶
.
R = (Q
0
t
− P
0
y
)/(tP − yQ)
ty
P + Qy
0
= 0 m(ty)
ty
(3t
2
y + 2ty + y
3
) dt + (t
2
+ y
2
) dy = 0
y
0
= 2e
2t
+ y − 1
dt + (t/y − sin y) dy = 0
y dt + (2ty − e
−2y
) dy = 0
e
t
dt + (e
t
ctg t + 2y cosec y) dy
[4(t
3
/y
2
) + (3/y)] dt + 3[(t/y)
2
+ 4y] dy = 0
³
3t +
6
y
´
+
³
t
2
y
+ 3
y
t
´
dy
dt
= 0
m(t, y) = [ty(2 t + y)]
−1
,
(3ty + y
2
) + (t
2
+ ty)y
0
= 0.
r(t)
S
0
= r(t)S
c(t)
S
0
(t) = r(t)S − c(t). (3.38)
r(t) = r = const, c(t) = c = const > 0. (3.39)
S
∗
= c/r
S > .S
∗
S
0 S
∗
S < S
∗
c >
rS
T S(T ) = 0
c > rS
0
S
0
S(0) = S
0
S(t) = S
∗
+ (S
0
− S
∗
)e
rt
, t > 0. (3.40)