Question

考研高数：微分方程。希望给出较为详细的分析思路和解答过程！ 回答的好，会再加分！

Answer 1

1
y''-xy'-y=1
y''-(xy'+y+1)=0
(y+1)''=y'' y'=(y+1)'
(y+1)''-(x(y+1)'+(y+1))=0
(y+1)''=(xy+x)'
设u=y+1
u''=(xu)'
du'/dx=d(xu)/dx
d(xu)=du'
xu+C0=u'
x(y+1)+C0=(y+1)'
xy+x+C0=(y+1)'
(y+1)'|x=0 = y'|x=0=0
C0=0
xu=u'
du/u=xdx
lnu=x^2/2+C1
u=Ce^(x^2/2)
y+1=Ce^(x^2/2)
y(0)=0
C=1
特解y=e^(x^2/2)-1

2
如果方程是(x^2-1)dy + (2xy-cosx)dx=0
x^2dy+ydx^2=d(sinx+y)
d(x^2y)=d(sinx+y)
x^2y=sinx+y+C0
y(0)=1
0=1+C0
C0=-1
特解 x^2y=sinx+y-1

追问

请问第一题您的思路是~~？

追答

(xy)'=x'y+y （y+1)'=y' (y+1)''=y''

然后xu+C0=u' 应用了题设y'(0)=0 找到C0常数，
x(y+1)+C0=(y+1)'
xy+x+C0=(y+1)'
(y+1)'|x=0 = y'|x=0=0
C0=0

追问

那第三题？

追答

y'+sin(x+y)=sin(x-y)/2
y'=sin(x-y)/2-sin(x+y)/2
=-2sin(y/2)cos(x/2)
y'=dy/dx=d(y/2)/d(x/2)
d(y/2)/d(x/2)=-2sin(y/2)cos(x/2)
d(y/2)/sin(y/2)=-2cos(x/2)d(x/2)
dcos(y/2)/(1-cosy/2)(1+cosy/2)=-2sin(x/2)
(1/2)ln|(1+cos(y/2)/(1-cos(y/2))|=-2sin(x/2)+C
ln|(1+cos(y/2)/sin(y/2)|=-2sin(x/2)+C

Answer 2

我只会用matlab算了，不过应付考研不行