TITLE: A third order nonlinear ordinary differential equation.
QUESTION [11 upvotes]: How can we solve $$y'''(t)+a(y''(t))^2+b(y'(t))^3=0$$
Could one make some kind of least common denominator argument to decide possible substitutions? Since the chain rule will come into play, I suppose a substitution both for the variable $t$ and the function $y$ could be useful. Possibly some powers of them?

REPLY [3 votes]: As a more general solution, if you have an equation of the form 
$$ x''(t) + a(x(t))x'(t)^2+b(x(t)) = 0 $$
then you can make the substitution $f(x) = x'(t)^2$ to arrive at the equation
$$ \frac{1}{2}f'(x) + a(x)f(x)+b(x)= 0 $$
Letting $\mu(x) = \exp\left[\int a(x) dx\right]$, we can solve for $f(x)$:
$$ f(x) = \mu(x)^{-1}\left(C_1-2\int\mu(x)b(x)dx\right) $$
which can be substituted back for $x(t)$:
$$ x' = \mu(x)^{-1/2}\left(C_1-2\int\mu(x)b(x)dx\right)^{1/2} $$
and solved implicitly:
$$ C_2 + t - \int \left[ \mu(x)\left(C_1-2\int\mu(x)b(x)dx\right)^{-1} \right]^{1/2} dx = 0 $$