How to solve this differential equation? $y''-\frac{y'}{x}=4x^{2}y$ How to solve  $y''-\displaystyle\frac{y'}{x}=4x^{2}y$ ?  
I know that the solution of this equation is: $y = e^{x^{2}}$, but I cannot resolve.
First I thought that $z=y'$ could be, but was not.
 A: Note that we have $xy'' - y' = 4x^3y$. The LHS in some sense is dimensional consistent and looks something similar to a quotient rule provided we divide by $x^2$. So dividing by $x^2$ and doing algebraic manipulations we get $(\frac{1}{x}y')' = 4xy$. Now this looks familiar to some extent.
Rewriting, we get $(\frac{y'}{2x})' = 2xy$.
Now let $\frac{y'}{2x} = z(x)$. Plug this in and simplify to get $z' = \frac{y'}{z}y$
(Replace $2x$ by $\frac{y'}{z}$).
So we have $z^2 = y^2 + c$.
Thus we have now converted a second order differential equation in terms of first order differential equation, viz,
$\frac{1}{2x}\frac{dy}{dx} = \pm \sqrt{y^2 + c}$.
where $c$ is a constant.
(You could plug this in and check that this satisfies the second order differential equation.)
We now need other conditions (boundary/ initial conditions) to completely solve the problem i.e. to eliminate $c$ and other constant which will arise after solving the first order differential equation to get $y(x) = \exp(x^2)$.
(Note that taking $c =0 $ we get a simple ode and the solution to which is $y(x) = y(0) \exp(\pm x^2)$).
$\textbf{EDIT:}$
The first order ODE can be solved as follows:
$\textbf{CASE 1:}$
Let $c > 0$, then let $c = a^2$
Rearranging, we get 
$\frac{dy}{\sqrt{y^2 + a^2}} = \pm d(x^2)$
$y = a \tan(\theta)$, we get $dy = a \sec^2(\theta) d\theta$.
Hence, the ode now becomes,
$\sec(\theta) d\theta = \pm d(x^2)$
$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.
$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$
$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$
Substitute for $\theta$ in terms of $y$ to get,
$\frac{y}{a} \pm \sqrt{1+(\frac{y}{a})^2} = K \exp(\pm x^2)$
$\textbf{CASE 2:}$
Let $c > 0$, then let $c = -a^2$.
Rearranging, we get 
$\frac{dy}{\sqrt{y^2 - a^2}} = \pm d(x^2)$
$y = a \sec(\theta)$, we get $dy = a \sec(\theta) \tan(\theta) d\theta$.
Hence, the ode now becomes,
$\sec(\theta) d\theta = \pm d(x^2)$
$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.
$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$
$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$
Substitute for $\theta$ in terms of $y$ to get,
$\frac{y}{a} \pm \sqrt{(\frac{y}{a})^2 - 1} = K \exp(\pm x^2)$
$\textbf{CASE 3:}$
Let $c = 0$.
The equation, we have now is $\frac{dy}{dx} = \pm 2xy$.
Solving, we get $y(x) = K \exp(\pm x^2)$.
A: $\rm\quad\quad\quad \ y'\ =\ f\ y $  
$\rm\quad\Rightarrow\ y''\ =\ f\:'\ y + f^{\:2}\ y $  
$\rm\quad\displaystyle\Rightarrow\ y''\ = \frac{f\:'}f\ y' + f^{\:2}\ y $
So $\rm\quad\displaystyle y''\ = \ \frac{1}x\ y' + 4x^2\ y\ \ \Rightarrow\ \ f\ =\ \pm 2\:x$
and $\rm\ y'\ =\ \pm 2\:x\ y\ \ \Rightarrow\ \ y\ =\ c\ e^{\pm x^2}$
A: $y''-\dfrac{y'}{x}=4x^2y$
$x\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}-4x^3y=0$
which belongs to a second order linear ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0208.pdf
From there, we get the hints of let $t=x^n$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=nx^{n-1}\dfrac{dy}{dt}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(nx^{n-1}\dfrac{dy}{dt}\right)=nx^{n-1}\dfrac{d}{dx}\left(\dfrac{dy}{dt}\right)+n(n-1)x^{n-2}\dfrac{dy}{dt}=nx^{n-1}\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)\dfrac{dt}{dx}+n(n-1)x^{n-2}\dfrac{dy}{dt}=nx^{n-1}\dfrac{d^2y}{dt^2}nx^{n-1}+n(n-1)x^{n-2}\dfrac{dy}{dt}=n^2x^{2n-2}\dfrac{d^2y}{dt^2}+n(n-1)x^{n-2}\dfrac{dy}{dt}$
$\therefore x\left(n^2x^{2n-2}\dfrac{d^2y}{dt^2}+n(n-1)x^{n-2}\dfrac{dy}{dt}\right)-nx^{n-1}\dfrac{dy}{dt}-4x^3y=0$
$n^2x^{2n-1}\dfrac{d^2y}{dt^2}+n(n-1)x^{n-1}\dfrac{dy}{dt}-nx^{n-1}\dfrac{dy}{dt}-4x^3y=0$
$n^2x^{2n-1}\dfrac{d^2y}{dt^2}+n(n-2)x^{n-1}\dfrac{dy}{dt}-4x^3y=0$
$n^2x^{2n-4}\dfrac{d^2y}{dt^2}+n(n-2)x^{n-4}\dfrac{dy}{dt}-4y=0$
$n^2t^{\frac{2n-4}{n}}\dfrac{d^2y}{dt^2}+n(n-2)t^{\frac{n-4}{n}}\dfrac{dy}{dt}-4y=0$
The ODE has the simplest form when we choose $n=2$.
The ODE becomes
$4\dfrac{d^2y}{dt^2}-4y=0$
$\dfrac{d^2y}{dt^2}-y=0$
The auxiliary equation is
$\lambda^2-1=0$
$\lambda=+-1$
$\therefore y=C_1e^t+C_2e^{-t}$
$y=C_1e^{x^2}+C_2e^{-x^2}$
