I'm having trouble solving this equation
$y'' + (e^{2y} + 1)(y')^3 = 0$
I did the substitution: $y' = v(y)$, leading to the following equation:
$v'v + (e^{2y} + 1)v^3 = 0$
I see that $v = 0$ is a solution so I divide out, leaving me with:
$v' + (e^{2y} + 1)v^2 = 0$
$v' = - (e^{2y} + 1)v^2$
$dy = - (e^{2y} + 1)v^2 dv$
Is this the right direction to proceed? The integral on the right seems intimidating, so I'm wondering if there's a better way.
If you interchange dependent and independent variables, you get a much nicer equation: $$ x''(y) = e^{2y} + 1$$ which can be solved by integrating twice.
You're fine as far as: $v' = - (e^{2y} + 1)v^2$
This is $\frac {dv}{dy} = - (e^{2y} + 1)v^2$
Your separation of variables has gone wrong: you should get:
$\int \frac 1 {v^2}{dv} = - \int (e^{2y} + 1) dy$
$- \frac 1 {v} = - \frac 12 e^{2y} -y +c$
$v = \frac 2 {e^{2y} +2y +c}$
Recall that $v=\frac {dy}{dx}$
$\frac {dy}{dx}= \frac 2 {e^{2y} +2y +c}$
$\int( {e^{2y} +2y +c})dy=\int2dx$
$\frac 12 e^{2y} +y^2+cy+d=2x$
$\frac 14 e^{2y} +\frac 12 y^2+cy+d=x$
To me, the most impressive approach is in Robert Israel's answer.
It is based on the fact (see here, here or here) that $$\frac{d^2x}{dy^2}=-\frac {\frac{d^2y}{dx^2} }{\left(\frac{dy}{dx}\right)^3}$$ that is to say that $$\frac{d^2y}{dx^2}=-\frac{d^2x}{dy^2}\left(\frac{dy}{dx}\right)^3$$ So, the differential equation
$$\frac{d^2y}{dx^2} + (e^{2y} + 1)\left(\frac{dy}{dx}\right)^3 = 0$$ just write $$-\frac{d^2x}{dy^2}\left(\frac{dy}{dx}\right)^3+ (e^{2y} + 1)\left(\frac{dy}{dx}\right)^3 = 0$$ Assuming $\frac{dy}{dx}\neq 0$ (that is to say $y\neq c$ which is a particular solution for the given differential equation), then $$\frac{d^2x}{dy^2}=e^{2y}+1\implies \frac{dx}{dy}=\frac{e^{2 y}}{2}+y+c_1\implies x=\frac{e^{2 y}}{4}+\frac{y^2}{2}+c_1y+c_2$$
All credit should be given to Robert Israel.
