(1-x^2)y''-xy'=2
y''-x/(1-x^2)y'=2/(1-x^2)
令u(x)=e^(∫-x/(1-x^2) dx)
u=e^(ln(x^2-1)/2)=(x^2-1)*sqrt(e)
由于
d(uy')/dx=u'y'+uy''=uy''-u*(x/(1-x^2))y'=2u
uy'=∫2u dx
uy'=sqrt(e)[x^3/3-x]+c
y'=[sqrt(e)[x^3/3-x]+c]/((x^2-1)*sqrt(e))\
整理,求積分得
y=ln[2 (x + Sqrt[-1 + x^2])] (-ln[2 (x + Sqrt[-1 + x^2])] + c1]) +c2