常微分习题5_2答案

答案5.2
1．解： 解曲线是：t x y t
x x sin cos 00==
相平面的轨线是： 2022x y x =+2．解： 由方程组1）,,cy x y
x dy dx == 过点（1，2）x y 2=∴，由非自治系统 101−−==−−t e c x t x dt
dx t t e c y y dt
dy 2== 过点（1，2） t t t t e y t e
t x +−+−=−−+=002,1)2(0 即 1ln 2ln )2
1(00−−+−+=y t y t x 3．证明：0000yx xy cy x cy
x y x dy dx ===∴=， 当时，从出发的解与初始时刻无关。

其原因是由于右端有相同的因子，可以通过时间变换将其变化为自治系统。

0t t =),(00y x 0t
4．证明: 1) 性质4:
设有两条轨线 及有 )(),(t y y t x x ==0,)
(),(0>∃∃+=+=T t T t y y T t x x )()(),()(0000T t y t y y T t x t x x +==+==
∴ 两条轨线交于同一点 ,由性质2可知. 由解的存在唯一性定理. ))(),((00t y t x 过相平面上一点系统有且仅有一条轨线通过.
∀),(00y x )()()
()(T t y t y T t x t x +=+=∴
2) 性质5：(反证法)
若系统出发于∀非奇点的轨线在有限时间内达到某奇点, 即有一轨线过点, 而系统又),(∗∗y x ),(∗∗y x ∃一条常轨线
,,∗∗==y y x x 由解的唯一性定理, 两条轨线不可能交于同一点.
∃∗∗==y y x x , ∴ 性质5得证.
5解: 由方程可得, 该方程组有4个平衡点
⎟⎟⎠
⎞⎜⎜⎝⎛−−−−12121121122122121122
,
)0,(),0()0,0(0βααββαβαβααβαβr r r r C r B r A 解轨线的走向请参阅例5.2.3和下面的图形
restart:with(DEtools):
ODE1 :=[diff(x(t),t)=x(t)*(3-2*x(t)-y(t)),
diff(y(t),t)=y(t)*(3-x(t)-2*y(t))];
DEplot( ODE1, [x(t),y(t)],t=-100..100,
[[x(0)=0,y(0)=1],[x(0)=0,y(0)=5],
[x(0)=0.1,y(0)=0.6],[x(0)=0.6,y(0)=0.1],
[x(0)=5,y(0)=0],[x(0)=2,y(0)=1],[x(0)=0.1,y(0)=0.1],
[x(0)=5,y(0)=5], [x(0)=1.5,y(0)=5],[x(0)=0.1,y(0)=0.2],
[x(0)=0.2,y(0)=0.1],[x(0)=1,y(0)=0],[x(0)=5,y(0)=3],
[x(0)=5,y(0)=0.5],[x(0)=0.5,y(0)=5],[x(0)=3,y(0)=5]], x=0..5,y=0..5,stepsize=0.05, dirgrid=[21,21],
color=red, linecolor=blue,axes=BOXED,
title="Phase portrait of the Linear System",
arrows=SLIM);
:= ODE1⎡⎣⎢⎢⎤⎦
⎥⎥, = ∂∂t ()x t ()x t () − − 32()x t ()y t = ∂∂t ()y t ()y t () − − 3()x t 2()y t
6．对不同的参数值所得到的解轨线如下
restart:with(DEtools): mu:=0.1; a:=4;
ODE1 :=[diff(x(t),t)=y(t),
diff(y(t),t)=-x(t)-mu*y(t)*(x(t)^2-1)];
DEplot( ODE1, [x(t),y(t)],t=-100..100,
[[x(0)=0.1,y(0)=0.1],[x(0)=0,y(0)=0.8*a],
[x(0)=0,y(0)=-0.8*a],
[x(0)=-0.8*a,y(0)=0],[x(0)=0.8*a,y(0)=0]],
x=-a..a,y=-a..a,stepsize=0.05, dirgrid=[21,21], color=red, linecolor=blue,axes=BOXED,
title="Phase portrait of the Nonlinear System", arrows=SLIM);
:= μ.1
:= a 4
:= ODE1⎡⎣⎢⎢⎤⎦
⎥⎥, = ∂∂t ()x t ()y t = ∂∂t ()y t − − ()x t .1()y t () − ()x t 21
restart:with(DEtools): mu:=0.5; a:=4;
ODE1 :=[diff(x(t),t)=y(t),
diff(y(t),t)=-x(t)-mu*y(t)*(x(t)^2-1)];
DEplot( ODE1, [x(t),y(t)],t=-100..100,
[[x(0)=0.1,y(0)=0.1],[x(0)=0,y(0)=0.8*a],
[x(0)=0,y(0)=-0.8*a],
[x(0)=-0.8*a,y(0)=0],[x(0)=0.8*a,y(0)=0]],
x=-a..a,y=-a..a,stepsize=0.05, dirgrid=[21,21], color=red, linecolor=blue,axes=BOXED,
title="Phase portrait of the Nonlinear System", arrows=SLIM);
:= μ.5
:= a 4
:= ODE1⎡⎣⎢⎢⎤⎦
⎥⎥, = ∂()x t ()y t = ∂()y t − − ()x t .5()y t () − ()x t 21
> restart:with(DEtools): mu:=1; a:=4;
ODE1 :=[diff(x(t),t)=y(t),
diff(y(t),t)=-x(t)-mu*y(t)*(x(t)^2-1)];
DEplot( ODE1, [x(t),y(t)],t=-100..100,
[[x(0)=0.1,y(0)=0.1],[x(0)=0,y(0)=0.8*a],
[x(0)=0,y(0)=-0.8*a],
[x(0)=-0.8*a,y(0)=0],[x(0)=0.8*a,y(0)=0]],
x=-a..a,y=-a..a,stepsize=0.05, dirgrid=[21,21], color=red, linecolor=blue,axes=BOXED,
title="Phase portrait of the Nonlinear System", arrows=SLIM);
:= μ1
:= a 4
:= ODE1⎡⎣⎢⎢⎤⎦
⎥⎥, = ∂∂t ()x t ()y t = ∂∂t ()y t − − ()x t ()y t () − ()x t 21
> restart:with(DEtools): mu:=2; a:=4;
ODE1 :=[diff(x(t),t)=y(t),
diff(y(t),t)=-x(t)-mu*y(t)*(x(t)^2-1)];
DEplot( ODE1, [x(t),y(t)],t=-100..100,
[[x(0)=0.1,y(0)=0.1],[x(0)=0,y(0)=0.8*a],
[x(0)=0,y(0)=-0.8*a],
[x(0)=-0.8*a,y(0)=0],[x(0)=0.8*a,y(0)=0]],
x=-a..a,y=-a..a,stepsize=0.05, dirgrid=[21,21], color=red, linecolor=blue,axes=BOXED,
title="Phase portrait of the Nonlinear System", arrows=SLIM);
:= μ2
:= a 4
:= ODE1⎡⎣⎢⎢⎤⎦
⎥⎥, = ∂∂t ()x t ()y t = ∂∂t ()y t − − ()x t 2()y t () − ()x t 21。