Solve: $\frac{d^2x}{dt^2}+\Bigl(\frac{d^2y}{dt^2}\Bigr)^2=0$ We have the following coupled Diferential equation:
$\dfrac{d^2x}{dt^2}=-\left(\dfrac{y}{x}\right)^2$
$\dfrac{d^2y}{dt^2}=\dfrac{y}{x}$
Then find the solution $x$ and $y$ in terms of $t$ .
What we have that
$\dfrac{d^2x}{dt^2}+\left(\dfrac{d^2y}{dt^2}\right)^2=0$
Now how to solve this equation. Please help. Thanks in advance!
 A: $\begin{cases}\dfrac{d^2x}{dt^2}=-\left(\dfrac{y}{x}\right)^2~......(1)\\\dfrac{d^2y}{dt^2}=\dfrac{y}{x}............(2)\end{cases}$
From $(2)$, $x=\dfrac{y}{\dfrac{d^2y}{dt^2}}......(3)$
Put $(3)$ into $(1)$ :
$\dfrac{d^2}{dt^2}\left(\dfrac{y}{\dfrac{d^2y}{dt^2}}\right)=-\dfrac{y^2}{\dfrac{y^2}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}}$
$\dfrac{d}{dt}\left(\dfrac{\dfrac{dy}{dt}}{\dfrac{d^2y}{dt^2}}-\dfrac{y\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}\right)=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^2$
$\dfrac{\dfrac{d^2y}{dt^2}}{\dfrac{d^2y}{dt^2}}-\dfrac{\dfrac{dy}{dt}\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}-\dfrac{\dfrac{dy}{dt}\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}-\dfrac{y\dfrac{d^4y}{dt^4}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}+\dfrac{2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^3}=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^2$
$1-\dfrac{y\dfrac{d^4y}{dt^4}+2\dfrac{dy}{dt}\dfrac{d^3y}{dt^3}}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^2}+\dfrac{2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2}{\biggl(\dfrac{d^2y}{dt^2}\biggr)^3}=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^2$
$\biggl(\dfrac{d^2y}{dt^2}\biggr)^3-y\dfrac{d^2y}{dt^2}\dfrac{d^4y}{dt^4}-2\dfrac{dy}{dt}\dfrac{d^2y}{dt^2}\dfrac{d^3y}{dt^3}+2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2=-\biggl(\dfrac{d^2y}{dt^2}\biggr)^5$
$y\dfrac{d^2y}{dt^2}\dfrac{d^4y}{dt^4}+2\dfrac{dy}{dt}\dfrac{d^2y}{dt^2}\dfrac{d^3y}{dt^3}-2y\biggl(\dfrac{d^3y}{dt^3}\biggr)^2-\biggl(\dfrac{d^2y}{dt^2}\biggr)^5-\biggl(\dfrac{d^2y}{dt^2}\biggr)^3=0$
