Differentiation with respect to two different variables If $\ x=e^t$ 
Show that (done)$\ x\frac{dy}{dx}=\frac{dy}{dt}$ and (need help on this) $$\ x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}$$
Use the results to reduce the differential equation to a differential equation in y and t. (No help required on this)
$$\ x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-4y=16$$
 A: $$\ x\frac{dy}{dx}=\frac{dy}{dt}\\ 
\implies \frac{d\ (x\frac{dy}{dx})}{dx} =^{\text{[1]}}  \frac{dy}{dx} + x\frac{d^2y}{dx^2} =^{\text{[2]}} \frac{dy}{xdt} + x\frac{d^2y}{dx^2}=^{\text{[3]}}\frac{d\frac{dy}{dt}}{dx}=^{\text{[4]}}{\frac{d^2y}{dxdt}} \implies \\ 
\frac{dy}{xdt} + x\frac{d^2y}{dx^2}=^{\text{[5]}}{\frac{d^2y}{xdt^2}} \\ 
\implies x^2\frac{d^2y}{dx^2} =^{\text{[6]}} {\frac{d^2y}{dt^2}}-\frac{dy}{dt}$$
$[1]$: because of product rule of derivative.
$[2],[5]$: because of assumption: $\ x\frac{dy}{dx}=\frac{dy}{dt}$
$[3]$: because of right hand side of the first formula.
$[4]$: rewrite the derivative.
$[6]$: multiply two side with $x$.
A: Let $y=y(x)$ then
$$\frac{d^2y}{dt^2}=\frac{d}{dt}\left(x\frac{dy}{dx}\right)=\frac{d}{dt}\left(e^t\frac{dy}{dx}(e^t)\right)=
e^t\frac{dy}{dx}(e^t)+\frac{d^2y}{dx^2}(e^t)\cdot\frac{dx}{dt}\cdot e^t\\=
x\frac{dy}{dx}+\frac{d^2y}{dx^2}\cdot x^2=\frac{dy}{dt}+\frac{d^2y}{dx^2}\cdot x^2$$
which implies $x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}.$
Then your ODE becomes $\frac{d^2y}{dt^2}-4y=16$
which has constant coefficients (and it is easier to solve!).
A: $$x^2\frac{d^2y}{dx^2}=x^2\frac{d}{dx}(\frac{dy}{dx})=x^2\frac{d}{dx}(\frac{1}{x}\frac{dy}{dt})=x^2(-\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x}\frac{d}{\frac{dx}{dt}dt}(\frac{dy}{dt}))=x^2(-\frac{1}{x^2}\frac{dy}{dt}+\frac{1}{x^2}\frac{d^2y}{dt^2})=-\frac{dy}{dt}+\frac{d^2y}{dt^2}$$
