what technique should I use to solve this $v'' e^{v'} -2v=0$ analytically.
Here is what I did:
$v''e^{v'}=2v$ integrate both sides wrt x to get $e^{v'}=v^2 +c$ then I took the natural logarithm to both sides, this gives:
$v'=\ln(v^2 +c)$ integrate again by parts this gives: $v=x \ln(v^2+c) - \int \frac{2v}{v^2 +c} dx = x \ln(v^2+c) - \ln(v^2 +c) + c_1$
Using shorthand you have \begin{equation} v''e^{v'} = 2 v. \end{equation} Multiply through by $v'$ to get \begin{equation} v''e^{v'}v' = 2 vv'. \end{equation} Integrate both sides: \begin{align*} \int v'' e^{v'} v' dt &= \int 2 v v' dt \\ \int e^{v'} v' dv' &= \int 2 v dv. \end{align*} Use integration by parts to compute the first integral on the left; you can do the final one directly. That’ll get you to a first order equation.