Solve $2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$ Solve differential equation
$$2\ddot{y}y - 3(\dot{y})^2  + 8x^2 = 0$$
I know that we have to use some smart substitution here, so that the equation becomes linear.
The only thing I came up with is a smart guessed particular solution: $y = x^2$. If we plug this function in, we get:
$$2\cdot2\cdot x^2 - 3(2x)^2 +8x^2 = 4x^2 - 12x^2 + 8x^2= 0$$

I made a mistake. The coefficients where different in the exam:
$$
\begin{cases}
    3\ddot{y}y + 3(\dot{y})^2 - 2x^2 = 0, \\
    y(0) = 1, \\
    \dot{y}(0) = 0.
\end{cases}
$$
Does it make the solution easier?

Hint:
Let $y=\dfrac{1}{u^2}$ ,
Then $y'=-\dfrac{2u'}{u^3}$
$y''=\dfrac{6(u')^2}{u^4}-\dfrac{2u''}{u^3}$
$\therefore\dfrac{2}{u^2}\left(\dfrac{6(u')^2}{u^4}-\dfrac{2u''}{u^3}\right)-3\left(-\dfrac{2u'}{u^3}\right)^2+8x^2=0$
$\dfrac{12(u')^2}{u^6}-\dfrac{4u''}{u^5}-\dfrac{12(u')^2}{u^6}=-8x^2$
$\dfrac{4u''}{u^5}=8x^2$
$u''=2x^2u^5$
This reduces to a special case of Emden-Fowler equation.
