I am trying to do this exercise:
Transform the equation: $y'' + \frac{2}{x}y'+y=0$ taking $x$ as a function and $t=xy \hspace{0.2cm}$ as the independent variable.
The answer is supposed to be: $\frac{d^2x}{dt^2}-t(\frac{dx}{dt})^3=0$.
All of the exercises I have done related to change of variable had in common that either $x$ or $y$ were the independent variable. I have no clue as to how to do this exercise. Any hints would be really helpful.
Thanks
For $y’’+\frac{2y’}{x}+y=0$, so $xy=-x\left(\frac{d}{dx}\left(\frac{dy}{dx}\right)+\frac{2}{x}\frac{dy}{dx}\right)$ $$\because t=xy, \therefore dt=xdy+ydx, \therefore \frac 1{\frac{dx}{dt}}=\frac{dt}{dx}=x\frac{dy}{dx}+y,\therefore \frac{dx}{dt}=\frac 1{x\frac{dy}{dx}+y}$$ $$\therefore d\frac{dx}{dt}=-\frac{d(x\frac{dy}{dx}+y)}{(x\frac{dy}{dx}+y)^2}=-\frac{xd\frac{dy}{dx}+dx\frac{dy}{dx}+dy}{(x\frac{dy}{dx}+y)^2}=-\frac{xd\frac{dy}{dx}+2dy}{(x\frac{dy}{dx}+y)^2} $$ $$\therefore \frac{d^2x}{dt^2}=\frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{dx}{dt}\cdot\frac{d}{dx}\left(\frac{dx}{dt}\right)=\frac{dx}{dt}\cdot\left( -\frac{x\frac{d}{dx}\left(\frac{dy}{dx}\right)+2\frac{dy}{dx}}{(x\frac{dy}{dx}+y)^2} \right)= \frac{dx}{dt}\frac {xy}{(x\frac{dy}{dx}+y)^2}=\frac t{(x\frac{dy}{dx}+y)^3} $$ So $\frac{d^2x}{dt^2}-t\left(\frac{dx}{dt}\right)^3=0$