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?

$$
\begin{cases}
    3y''y + 3(y')^2 - 2x^2 = 0, \\
    y(0) = 1, \\
    y'(0) = 0.
\end{cases}
$$
$$y''y+y'^2=(y'y)'\quad\to\quad 3(y'y)'=2x^2$$
$$3y'y=\frac{2}{3}x^3+c_1$$
$y'(0)=0\quad\to\quad c_1=0$
$$y'y=\frac{2}{9}x^3$$
$$2y'y=(y^2)'=\frac{4}{9}x^3$$
$$y^2=\frac{1}{9}x^4+c_2$$
$y(0)=1\quad\to\quad c_2=1$
$$y=\sqrt{\frac{1}{9}x^4+1}$$ 
