How to solve the homogeneous ODE
$$ \frac{dx}{dt} = \frac{2tx}{t^2 - x^2} $$
I’ve changed the variable such that $y=x/t$ and I came up with the following
$$ \frac{dy}{dt} = -\frac{y}{t} + \frac{2yt}{1-y^2} $$
Everything I have tried after this result, failed. Can someone give me a hint, how to find the solution?
Thank you!
Answering just to remove this from the unanswered questions list: as was noted in the comments, a substitution $x(t)=tv(t)$ transforms the ODE into
$$ v+v't=\frac{2v}{1-v^2} $$
which is equidimensional in $t$; i.e. it is invariant under a rescaling $t\rightarrow at$. Changing the independent variable to $t=e^{\tau}$ produces
$$ v+v'=\frac{2v}{1-v^2} $$
where the derivative is evaluated with respect to $\tau$. This is a first-order separable ODE:
$$ \begin{align} \left(\frac{1}{v}-\frac{2v}{1+v^2}\right) dv&=d\tau\\ \ln\left(\frac{|v|}{1+v^2}\right)&=\tau-c_1\\ \frac{|v|}{1+v^2}&=c_1e^{\tau} \end{align} $$
We can remove the absolute values bars on $v$ by letting $c_1\in\mathbb{R}$. Changing variables from $\tau$ back to $t$, and from $v$ back to $x$:
$$ \begin{align} \frac{v}{1+v^2}&=c_1t\\ \frac{t\cdot(vt)}{t^2+(vt)^2}&=c_1t\\ \frac{x}{t^2+x^2}&=c_1 \end{align} $$
Solving for $x(t)$, we obtain
$$ x(t)=\frac{c_1\pm\sqrt{c_1^2-4t^2}}{2} $$
Note that $x=0$ is also a solution to the ODE which we cannot obtain from any choice of $c_1$ in the above.