Consider the following second-order nonlinear ODE: $$y'' = e^y$$ The solution to this ODE is known precisely: $$y = \ln\left(\frac{1}{2}c_{1}\left(\tanh^2\left(\frac{1}{2}\sqrt{c_1(c_2+x^2)}\right)-1\right)\right)$$ Given this knowledge about the solution, is there any way to transform this ODE into a linear one using successive changes of variables?
A motivating example; consider the first-order analogue $y' = e^y$, which has the solution:
$$y = \ln\left(\frac{1}{c - x}\right)$$
If I wanted to transform this nonlinear ODE into a linear one, I might consider the following coordinate transformation, $y = \ln(z)$; doing so generates the following nonlinear (but algebraic) ODE for $z$:
$$z' = z^2$$
Following up with another change of variables $z = w^{-1}$, we now find a linear ODE for $w$:
$$w' = -1, \quad w = c - x$$
and the solution for $y$ can be found trivially by applying all the coordinate changes to the solution of the linear ODE.
In short, by observing that the solution is a composition of operations on a relatively simple function ($c - x$), we could apply those operations consecutively on the ODE to eventually get a linear ODE. Can we do that for $y'' = e^y$ too?
$$y''=e^y$$ First substitution : $\quad y'=u(y)$
$y''=u'(y)y'=u'(y)u(y)$ $$uu'=e^y$$ Second substitution :$\quad v(y)=u^2$
$v'=2uu'=2e^y$ $$v'(y)=2e^y\quad\text{is a linear ODE.}$$
To make it more obvious, let :
$X=y\quad$ and $\quad v(y)=Y(y)=Y(X)$. $$Y'(X)=2e^X$$
Solving (outside the scope of the question) :
This is a linear first order separable ODE.
$v(y)=2\int e^y dy=2e^y+C_1$
$u=\pm\sqrt{2e^y+C_1}=\frac{dy}{dx}$
$x=\pm\int \frac{dy}{\sqrt{2e^y+C_1}} = \pm \frac{2}{\sqrt{C_1}}\tanh^{-1}\left(\sqrt{1+\frac{2}{C_1}e^y}\right)+C_2$
$\pm\sqrt{1+\frac{2}{C_1}e^y}=\tanh\left(\frac{\sqrt{C_1}}{2}(x-C_2) \right)$
$e^y=\frac{C_1}{2}\left(\tanh^2\left(\frac{\sqrt{C_1}}{2}(x-C_2) \right)-1\right)$
$y=\ln\left(\frac{C_1}{2}\left(\tanh^2\left(\frac{\sqrt{C_1}}{2}(x-C_2) \right)-1\right)\right)$