Question

X^2 y''+4xy'+2y=0 cauchy euler equation

Answer 1

Let $y=x^r$, then

$x^2(r(r-1)x^{r-2})+4x(rx^{r-1})+2x^r=0$
$r(r-1)x^r+4rx^r+2x^r=0$
$r(r-1)+4r+2=0$
$r^2+3r+2=0$
$(r+2)(r+1)=0$
$r=-1,-2$

This admits a general solution of

$y=C_1x^{-1}+C_2x^{-2}$