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?
 A: Let $z=e^{ky}$. Then $z'=ky'e^{ky}$ and $$z''=ky''e^{ky}+k^2(y')^2 e^{ky}\text.$$
If we put $k=-\frac{3}{2}$ then $z''=-\frac{3}{2}y''e^{ky}+\frac{3}{2}^2(y')^2 e^{ky}$, or $\frac{2}{3}z''=-y''e^{ky}+\frac{3}{2}(y')^2 e^{ky}$, or $$-\frac{4}{3}z''=2y''e^{ky}-3(y')^2 e^{ky}\text{, or even}$$ $$-\frac{4z''}{3z}=2y''-3(y')^2\text.$$
Your original equation is now $$\frac{4z''}{3z}=8x^2\text,$$ which I hope will get you closer.
Indeed...
Given $z''=6x^2z$, expanding $z$ as a power series in $x$ yields $a_n=\frac{a_{n-2}}{(n+1)(n+2)}$ times a constant, which makes one think that $z$ must be something to do with $e^{x^2}$.
Put $$z=e^{ax^2+bx+c}$$ and differentiate twice, and you get something like $$z''=2az+4a^2x^2z+2abxz+b^2z\text,$$ which is enticing but has an embarrassing $xz$ term. Fortunately that is the only term in which $b$ isn't squared. So if you take $$z=e^{ax^2+bx+c}+e^{ax^2-bx+c}$$ and differentiate it twice, you will get rid of the unwanted $xz$ term and you will have constraints on $a$ and $b$ which make $z$ satisfy the differential equation.
General note: All this has been done on the back of one and a half envelopes, so do check it.
A: $$
\begin{cases}
    3y''y + 3(y')^2 - 2x^2 = 0, \\
    y(0) = 1, \\
    y'(0) = 0.
\end{cases}
$$
$$y''y+y'^2=(y'y)'\quad\to\quad 3(y'y)'=2x^2$$
$$3y'y=\frac{2}{3}x^3+c_1$$
$y'(0)=0\quad\to\quad c_1=0$
$$y'y=\frac{2}{9}x^3$$
$$2y'y=(y^2)'=\frac{4}{9}x^3$$
$$y^2=\frac{1}{9}x^4+c_2$$
$y(0)=1\quad\to\quad c_2=1$
$$y=\sqrt{\frac{1}{9}x^4+1}$$ 
A: 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.
