Answer to Question #149878 in Differential Equations for Nikhil

Question

Solve
x^2y''- 2xy'+2(1+x^2)y=0 , x>0

Answer 1

$x^2y''-2xy'+2(1+x^2)y=0, x>0\\$

Divide through by $x^2$

$y''-\frac{2}{x}y'+\frac{2(1+x^2)}{x^2}y=0\\$

For the equation of the form $y''+f(x)y'+g(x)y=F(x).$ The substitution $y=v.e^{- \int \frac{f(x)}{2}dx}$ is used.

So,

$y=v.e^{- \int \frac{-2}{2x}dx}\\ y=v.e^{\int\frac{1}{x}dx}\\ y=v.e^{\ln x}\\ y=vx\\ y'=v+xv'\\ y''=2v'+xv''$

Rewrite the equation

$(2v'+xv'')-\frac{2}{x}(v+xv')+\frac{2(1+x^2)}{x^2}vx=0\\ xv''+2v'-\frac{2v}{x}-2v'+\frac{2(1+x^2)}{x}v=0\\ xv''-\frac{2}{x}v+\frac{2(1+x^2)}{x}v=0\\ xv''+\frac{2}{x}v(-1+1+x^2)=0\\ xv''+\frac{2}{x}v(x^2)=0\\ xv''+2xv=0$

divide through by $x$

$v''+2v=0$ .

The associated auxilliary equation to this is ;

$m^2+2m=0\\ m(m+2)=0\\ m=0,m=-2$

So,

$v=A_1e^{0.x}+A_2e^{-2.x}\\ v=A_1+A_2e^{-2x}$

But $v=\frac{y}{x}$

$\frac{y}{x}=A_1+A_2e^{-2x}\\ y=x(A_1+A_2e^{-2x}) ,x>0$