<s>[INST] How do I solve this system of non-linear differential equations? How do I solve this system of differential equations?
\begin{equation}
\frac{d^2x}{dt^2}+2\frac{1}{y}\frac{dx}{dt}\frac{dy}{dt}=0 \quad\quad 
\frac{d^2y}{dt^2}+\frac{1}{y}
\left[\left(\frac{dx}{dt}\right)^2-\left(\frac{dy}{dt}\right)^2\right]=0
\end{equation}
I would also like to know if there is some resource that is just a catalogue of different types of differential equations and their solutions (without any proofs and such) just for practical use. I don't usually work with D.E.s and forgot everything. 
 [/INST] Rearranging the first equation and integrate 
\begin{align} \frac{\ddot x}{\dot x} + \frac{2\dot y}{y} &= 0 \\
\ln(\dot x) + 2\ln(y) &= \ln(a)
\end{align}
$$ \implies \dot x = \frac{a}{y^2} $$
Plugging this into the second equation
$$ \ddot y + \frac{1}{y}\left[\frac{a^2}{y^4} - (\dot y)^2\right] = 0 $$
This equation is autonomous, so we substitute $v = \dot y$ and obtain a first-order ODE
$$ v\frac{dv}{dy} - \frac{v^2}{y} + \frac{a^2}{y^5} = 0 $$
which is a Bernoulli equation, so another substitution $z = v^2$ gives
$$ \frac{dz}{dy} - \frac{2}{y}z + \frac{2a^2}{y^5} = 0 $$
which has the solution
$$ z = b^2y^2 - \frac{a^2}{3y^4} $$
Going back to $\dot y$ and rearrange
$$ \dot y = \frac{\sqrt{3b^2y^6 - a^2}}{\sqrt{3}y^2} $$
$$ \frac{y^2\dot y}{\sqrt{y^6 - \dfrac{a^2}{3b^2}}} = b $$
which can be integrated to give
$$ y = \left[\frac{a}{\sqrt{3}b}\cosh(3bt + c)\right]^{1/3} $$
and
$$ \dot{x} = \left[\frac{3b^2a}{\cosh^2(3bt+c)}\right]^{1/3} $$
unfortunately $x$ does not have a nice form
Note: A special case occurs when $a=0$, then we have a different solution set
$$ x = c_1, \quad y = c_2e^{bt} $$
</s>