Can someone check my working for this question

By making the substitution $\displaystyle x=e^t$, reduce the following linear differential equation to one with constant coefficients, and hence solve it:

$\displaystyle x^2\frac{d^2y}{dx^2}-4x\frac{dy}{dx}+6y=x^5 \ \ \ \ \ \ , x>0$

This is what I've done so far:

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}$

$\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot e^{-t}$

$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\left(e^{-t}\,\frac{dy}{dt}\right)$

$\displaystyle = e^{-t}\,\frac{d}{dx}\left(\frac{dy}{dt}\right) + \frac{d}{dx}(e^{-t})\,\frac{dy}{dt}$

$\displaystyle = e^{-t}\,\frac{d}{dt}\left(\frac{dy}{dt}\right)\,\frac{ dt}{dx} + \frac{d}{dt}(e^{-t})\,\frac{dt}{dx}\,\frac{dy}{dt}$

$\displaystyle = e^{-2t}\,\frac{d^2y}{dt^2} - e^{-2t}\,\frac{dy}{dt}$

$\displaystyle e^{2t}\left(e^{-2t}\,\frac{d^2y}{dt^2} - e^{-2t}\,\frac{dy}{dt}\right)-4e^t\left(e^{-t}\frac{dy}{dt}\right)+6y=e^{5t}$

$\displaystyle \frac{d^2y}{dt^2}-5\frac{dy}{dt}+6y=e^{5t}$

Then would you solve that like a normal ODE