Answer to Question #209029 in Differential Equations for Shrikant Bharti

Question

Find regular singular point of (x^2+1)y"+y'+y=0

Answer 1

$(x^2+1)y''+y'+y=0$

$a_2(x)=(x^2+1)=0$

$\implies$ $x^2=-1$

Write the differential equation in standard form to get

"y"+{1\\over (x^2+1)}y'+{1\\over (x^2+1)}y=0"

$\implies p(x) ={1\over (x^2+1)}$ and $q(x)={1\over (x^2+1)}$

Point $x^2=-1$

Both appear to the first power of p(x) and q(x) hence it gives to points which are both regular singular points.the points are

$x=\sqrt-1$ and $x=-\sqrt-1$