Answer
\[y = c x^2lnx\]
Work Step by Step
$x^2y''-3xy'+4y=0$ ______($\alpha$)
$y=cx^2lnx$ _____(1)
Differentiate (1) with respect to $x$
\[y'=c[2xlnx+\frac{x^2}{x}]\]
$y'=c[2xlnx+x]$ _____(2)
Differentiate (2) with respect to $x$
$y'' = c[2lnx+\frac{2x}{x}+1]$
$y'' = c[2lnx+3]$. _______(3)
$x^2y''=c[2x^2lnx+3x^2]$ ______(4)
$-3xy'= c[-6x^2lnx-3x^2]$ ____(5)
$4y= 4cx^2lnx$ _____(6)
Adding (4),(5) and (6)
\[x^2y''-3xy'+4y=c[2x^2lnx+3x^2-6x^2lnx-3x^2+4x^2lnx]=0\]
Hence, $y=cx^2lnx$ is solution of ($\alpha$)
