求解常微分方程（1-x）y''+xy'-y=0,

Question

求解常微分方程
（1-x）y''+xy'-y=0,

Answer 1

(1-x)y''+xy'-y=0
(xy'-y)'=xy''
y''=(1/x) (xy'-y)'
[(1-x)/x](xy'-y)'+(xy'-y)=0
d(xy'-y)/(xy'-y)=xdx/(x-1)
ln|xy'-y|=x+ln|x-1|+lnC
(xy'-y)=C(x-1)e^x
xy'-y=C*(x-1)*e^x
xdy-ydx=C(x-1)e^xdx
d(y/x)=C(x-1)e^xdx/(x^2)
y/x=Ce^x/x +C1
y=Ce^x+C1x
∫(x-1)e^xdx/x^2=∫e^xdx/x-∫e^xdx/x^2
=∫de^x/x+∫e^xd(1/x)
=∫d(e^x/x)=(1/x)e^x+C1