A third order nonlinear ordinary differential equation. 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?
 A: $$y'''(t)+a(y''(t))^2+b(y'(t))^3=0$$
Substitute $z=y'$
$$z''(t)+a(z'(t))^2+bz^3=0$$
Substitute $p=z'$
$$\frac {dp}{dz}p+ap^2+bz^3=0$$
$$\frac 12(p^2)'+ap^2+bz^3=0$$
Finally substitute $w=p^2$
$$\frac 12w'+aw+bz^3=0$$
Bernouilli's equation
A: 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:
$$ C_2 + t - \int \left[ \mu(x)\left(C_1-2\int\mu(x)b(x)dx\right)^{-1} \right]^{1/2} dx = 0 $$
A: As Adrian suggests, let $y'=z$, to get the second order equation
$$
z''+a\,(z')^2+b\,z^3=0.
$$
Since the independent variable $t$ does not appear explicitly in the equation (I am assuming $a$ and $b$ are constants), we let
$$
z'=p,\quad z''=\frac{dp}{dt}=\frac{dp}{dz}\,\frac{dz}{dt}=p\,\frac{dp}{dz}.
$$
This gives the first order equation
$$
p\,\frac{dp}{dz}+a\,p^2+b\,z^3=0,
$$
which written as
$$
\frac{dp}{dz}=-a\,p-b\,z^3\,p^{-1}
$$
is a Bernoulli equation. To solve it, let $u=p^2$. This gives the linear equation
$$
\frac{du}{dz}=-a\,u-b\,z^3.
$$
I have not done the calculations, but my impression is that you will not be able to get an explicit solution in terms of elementary functions.
