How do we solve $y''=e^{2y}$? I tried to make the substitution $v=y'(x)$, which would ( hopefully ) make this into a separable ODE as per the following. $$y''=e^{2y}$$ by the chain rule, we have $v'=v'(y(x)) \cdot y'(x)=\frac{dv}{dy}\frac{dy}{dx}$. If we make the substitution, then $$\frac{dv}{dy}\frac{dy}{dx}=e^{2y}$$
So we could cancel the differential terms $dy$? I wonder if we can really do so, and also a better substitution might be just $v=y(x)$. Am I on the right track here?
 A: Multiply both sides by $y'$
$$y'y''=y'e^{2y} \implies (y')^2 = e^{2y}\pm C_1^2$$
Then square root and separate variables
$$\frac{e^{-y}dy}{e^{-y}\sqrt{e^{2y}\pm C_1^2}}= \pm dx$$
$$ \begin{cases} \frac{\sinh^{-1}(C_1e^{-y})}{C_1} = \pm x + C_2 \\ \frac{\sin^{-1}(C_1 e^{-y})}{C_1} = \pm x + C_2 \\ e^{-y} = \pm x + C \end{cases}$$ depending on the sign of the $C_1^2$ term. Each of these is invertible to get
$$\begin{cases} y = -\log(\pm \sinh(C_1 x + C_2) ) + \log C_1 \\ y = -\log(\pm \sin(C_1 x + C_2)) + \log C_1 \\ y = -\log(\pm x + C) \end{cases}$$
Note the first and second sets of $C_2$s are not equivalent.
A: Another way to do it.
Switch vcariables
$$y''=e^{2y} \implies -\frac{x''}{[x']^3}=e^{2y}$$ Reduction of order $p=x'$ leads to
$$p'=-p^3 \,e^{2y} \implies p=\pm\frac{1}{\sqrt{e^{2 y}+ c_1}}$$ One more integration
$$x+c_2=\mp \frac{\tanh ^{-1}\left(\frac{\sqrt{e^{2 y}+c_1}}{\sqrt{c_1}}\right)}{\sqrt{c_1}}$$ which is easy to inverse.
A: You have
$$y^{\prime \prime}(x) y^{\prime}(x) = y^{\prime}(x) e^{2y}$$
hence
$$(y^{\prime}(x))^2 -(y_0^\prime)^2= e^{2y(x)} - e^{2y_0}$$
From there
$$\frac{y^\prime(x)}{\sqrt{e^{2y(x)} - e^{2y_0}+(y_0^\prime)^2}} = \pm 1$$ and
$$\int_{y_0}^{y(x)}\frac{dt}{\sqrt{e^{2t} - e^{2y_0}+(y_0^\prime)^2}}= \pm (x - x_0)$$
A: Observe that with $v\equiv\frac{dy}{dx}$ you have:
$$
\frac{d^2y}{dx^2}=\frac{dv}{dx}=\frac{dv}{dy}\frac{dy}{dx}=v\frac{dv}{dy}\tag1
$$
With this your equation transforms to:
$$
v\frac{dv}{dy}=e^{2y}\implies v^2=e^{2y}+C\implies\frac{dy}{dx}=\pm\sqrt{e^{2y}+C}.
$$
Can you take it from here?
