<s>[INST] 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.
 [/INST] 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 $y(x) = K \exp(\pm x^2)$.
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