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)`