Answer
$y=C_{1}x^2\cos(3\ln x)$ is solution of ($\alpha$).
Work Step by Step
$x^2y''-3xy'+13y=0$ _____($\alpha$)
$y=C_{1}x^2\cos(3\ln x)$ ____(1)
Differentiate (1) with respect to $x$
$y'=C_{1}[2x\cos(3\ln x)-\frac{3x^2}{x}\sin(3\ln x)]$
$y'=C_{1}x[2\cos(3\ln x)-3\sin(3\ln x)]$ ___(2)
Differentiate (2) with respect to $x$
$y''=C_{1}[2\cos(3\ln x)-3\sin(3\ln x)]+C_{1}x[\frac{-6}{x}\sin(3\ln x)-\frac{9}{x}\cos(3\ln x)]$
$y''=C_{1}[-7\cos(3\ln x)-9\sin(3\ln x)]$ __(3)
$x^2y''=C_{1}[-7x^2\cos(3\ln x)-9x^2\sin(3\ln x)]$ ___(4)
$-3xy'=C_{1}[-6x^2\cos(3\ln x)+9x^2\sin(3\ln x)]$ ____(5)
$13y=13C_{1}x^2\cos(3\ln x)$ ___(6)
Adding (4),(5) and (6)
$x^2y''-3xy'+13y=C_{1}[-13x^2\cos(3\ln x)-9x^2\sin(3\ln x)+13x^2\cos(3\ln x)+9x^2\sin(3\ln x)]=0$
Hence, $y=C_{1}x^2\cos(3\ln x)$ is solution of ($\alpha$).
