How do I solve this system of non-linear differential equations?

Question

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.

Answer 1

One family of solutions is $x = c$, $y = a e^{bt}$ for constants $a,b,c$. Another is $x = a (t+b)^{1/3}$, $y = \pm a (t+b)^{1/3}/\sqrt{3}$ for constants $a,b$.

Answer 2

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

$$ x = c_1, \quad y = c_2e^{bt} $$