《常微分方程》_(方道元_著)_课后习题答案__浙江大学出版社

1. 1000 / -20000 / 2

h a=°2oooom/h2 v0=1000m/h, T

8>>><>>>:

dh

=v0+at.

h(T)=0

dh

|t=T

=0

122

h=v0t+1at+C, C v0T+aT+C=0,

0 T=°v h(0)=C=25m.

2. V A V1 / A V2 / V1+V2 / 2000 A 5m0 2000 A

m A

P(t)=e°

A P(t),

8

<VP0(t)+P(t)(V1+V2)=:P(0)=5m

V+VmV1

t

[5m0+

V+VmVt(e12

3.

(x,y) v0 v1

8>>><>>>:

dy°1)].

=v0=v1

x(0)=0,y(0)=0

4. 200kg º 150kg º

∏∫c ∏∫s ∫c ∫s ∏

1

mc,ms Vc,Vs,

8

<mcg°∏Vc=mcdVc+ΩVcg

:mg°∏V=mdV+ΩVg

s

s

ss

M = s

8

<mcg°ΩVcg=Mc:mg°ΩVg=M

s

s

s

m4

= s

8

<Vc(t)=:V(t)=

s

Mm(1Mm(1

°e°ct)°e°st)

∏

∏

1.

(1)

dydydy

(3)x2°2x+y=2xsinx;dyxdy(5)y°e+3xy=0;

3

=4x3°ysinx;

2

(2)(6)

y

dy(4)+cosy+4x=0;

dy+3°6y=0.

dy2°()+2xy;

d3y

(1) (2) (3) (4) (5) (6)

2. (1)y=1+x2,

dy(3)y=C1e2x+C2e°2x,(4)y=cxex,

d2y2

dy1

(2)y=°,x2°x2y2°xy°1=0;

dy

=y2°(x2+1)y+2x;

°4y=0( C1 C2 );

dy2dy

(5)y=ecx,()°y=0( c );

dy

°2+y=0( c );

2

(6)y=

C2<x<+1,

(1) (2) (3) (4) (5) (6) 3.

(1) y=Cx+x2 (2) xy=C (3) (4) y=9C+(1)y00=2 (2)y+xy0=0

(3)y000+(y0)2y000=3y0(y00)2 (4)y000=0

1. (1)(2)

dy

d2yC2

x

8

(x°C1)2>°,°1<x<C1;>><>>>:

0,C1<x<C2,

(x°C1)2

,dy

=

p x

=y+sinx;°

y=1+x,x

ex+0y(t)dt;x4+y3;

y(0)=1;

(3)y=(4)

dy=

(6)(ylnx°2)ydx=xdy;

(5)2xydy°(2y2°x)dx=0;

dy

(7)3xy2+y3+x3=0;

(8)

dy

=

y.

3

(1) e°x

y=Cex°

C (2) e

r

ºy=(C+)

q1+xº

C C=1° y=

°R

1

sinx+cosx

dx

(3) x

y0=ex+y(x) y(0)=1 e°x

y=ex(C+x)

C C=1 y=ex(1+x) (4) 3y2 z=y3

dz3z=3x3+dxx

e°

R

3

dx

z=x3(C+3x)

C y3=x3(C+3x) (5) z=y2

dz2z=°1dxx

2

e

°

R

dx

z=x2(C+x°1

y2=x2(C+x°1) C

(6) y¥0 y=0 °y°2 z=y°1

dz2zlnx=°dxxx

e°

R

dx

y°1=x2(C+

4

1+2lnx

)2

C

(7) y=0 y=0 y0=°1y° e

R

1

x22

3y z

=y3

dx

Cx3y=°

x C

xy

d(3

)

=°x3

°xdy

(8) y¥0 y=0 ydx°ydy=0 xy2

°=Cy C 2. y='(x)

dy

+a(x)y∑0,(x∏0).dx

'(x)∑'(0)e

e

Rx

°

Rx

a(t)dt

,(x∏0).

a(s)ds

3. f(x) (°1,+1)

dy

°y=f(x)dx

(°1,+1) f(x) ! !

|f(x)|∑M e°x y=ex(C+

R

f(s)e°sds) C

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