Question

求微分方程的通解.(1-x^2)y"-xy'=2

Answer 1

(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,求积分得

y=ln[2 (x + Sqrt[-1 + x^2])] (-ln[2 (x + Sqrt[-1 + x^2])] + c1]) +c2