齐次微分方程怎么解？

先算对应的齐次方程的解.
y'+P(x)y=0
y'/y=-P(x)
lny=-∫P(x)dx+C
y=ke^(-∫P(x)dx)
下面用常数变易法求解原方程的解.
设k为u(x)
y=u(x)e^(-∫P(x)dx)
y'=u'(x)e^(-∫P(x)dx)-u(x)P(x)e^(-∫P(x)dx)
代入得：
Q(x)
=u'(x)e^(-∫P(x)dx)-u(x)P(x)e^(-∫P(x)dx)+u(x)P(x)e^(-∫P(x)dx)
u(x)=∫Q(x)e^(∫P(x)dx)+C
y=e^(-∫P(x)dx)(∫Q(x)e^(∫P(x)dx)+C)

y''+y=cosx+e^x
The aux. equation
p^2+1=0
p=i or -i
let
yg= Dcosx +Esinx
yp = x(Acosx+Bsinx) + Ce^x
yp'=(Acosx+Bsinx) +x(-Asinx+Bcosx) +Ce^x
yp''
=(-Asinx+Bcosx) +(-Asinx+Bcosx) +x(-Acosx-Bsinx) +Ce^x
=2(-Asinx+Bcosx) +x(-Acosx-Bsinx) +Ce^x
yp''+yp = cosx+e^x
2(-Asinx+Bcosx) +x(-Acosx-Bsinx) +Ce^x + x(Acosx+Bsinx) + Ce^x =cosx+e^x
2(-Asinx+Bcosx) + 2Ce^x =cosx+e^x
=>
-2A=0 , 2B=1 and 2C=1
A=0, B=1/2 and C=1/2
yp= (1/2)xsinx + (1/2)e^x
通解
y=yg+yp=Dcosx +Esinx +(1/2)xsinx + (1/2)e^x