x^2 y''+4xy'+2y=0 cauchy euler equation
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\(y=C_1x^{-1}+C_2x^{-2}\)
