# How to solve this differential equation? y''-\frac{y'}{x}=4x^{2}y

How to solve this differential equation? $y{}^{″}-\frac{{y}^{\prime }}{x}=4{x}^{2}y$
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Debbie Moore
Note that we have $xy\prime \prime -y\prime =4{x}^{3}y$. 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 $\left(\frac{1}{x}y\prime \right)\prime =4xy$. Now this looks familiar to some extent.
Rewriting, we get ${\left(\frac{{y}^{\prime }}{2x}\right)}^{\prime }=2xy$
Now let $\frac{{y}^{\prime }}{2x}=z\left(x\right)$. Plug this in and simplify to get ${z}^{\prime }=\frac{{y}^{\prime }}{z}y$
(Replace 2x by $\frac{{y}^{\prime }}{z}$)
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,
$\frac{1}{2x}\frac{dy}{dx}=±\sqrt{{y}^{2}+c}$ 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 $y\left(x\right)=\mathrm{exp}\left({x}^{2}\right)$
(Note that taking $c=0$ we get a simple ode and the solution to which is $y\left(x\right)=y\left(0\right)\mathrm{exp}\left(±{x}^{2}\right)\right)$.
EDIT:
The first order ODE can be solved as follows:
CASE 1:
Let $c>0$, then $c={a}^{2}$
Rearranging, we get
$\frac{dy}{\sqrt{{y}^{2}+{a}^{2}}}=±d\left({x}^{2}\right)$
$y=a\mathrm{tan}\left(\theta \right)$, we get $dy=a{\mathrm{sec}}^{2}\left(\theta \right)d\theta$
Hence, the ode now becomes,
$\mathrm{sec}\left(\theta \right)d\theta =±d\left({x}^{2}\right)$
$d\left(\mathrm{log}\left(\mathrm{sec}\left(\theta \right)+\mathrm{tan}\left(\theta \right)\right)\right)=±d\left({x}^{2}\right)$
$\mathrm{log}\left(\mathrm{sec}\left(\theta \right)+\mathrm{tan}\left(\theta \right)\right)\right)=±\left({x}^{2}+k\right)$
$\mathrm{sec}\left(\theta \right)+\mathrm{tan}\left(\theta \right)=\mathrm{exp}\left(±\left({x}^{2}+k\right)\right)$
Substitute for $\theta$ in terms of y to get,
$\frac{y}{a}±\sqrt{1+{\left(\frac{y}{a}\right)}^{2}}=K\mathrm{exp}\left(±{x}^{2}\right)$
CASE 2:
Let $c>0$, then let $c=-{a}^{2}$
Rearranging, we get
Cassandra Ramirez
${y}^{\prime }=fy$
$⇒y{}^{″}={f}^{\prime }y+{f}^{2}y$
$⇒y{}^{″}=\frac{{f}^{\prime }}{f}{y}^{\prime }+{f}^{2}y$
So $y{}^{″}=\frac{1}{x}{y}^{\prime }+4{x}^{2}y⇒f=±2x$
and ${y}^{\prime }=±2xy⇒y=c{e}^{±{x}^{2}}$

alenahelenash

${y}^{″}-\frac{{y}^{\prime }}{x}=4{x}^{2}y$
$x\frac{{d}^{2}y}{d{x}^{2}}-\frac{dy}{dx}-4{x}^{3}y=0$
From there, we get the hints of left $t={x}^{n}$,
Then $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=n{x}^{n-1}\frac{dy}{dt}$
$\frac{{d}^{2}}{d{x}^{2}}=\frac{d}{dx}\left(n{x}^{n-1}\frac{dy}{dt}\right)=n{x}^{n-1}\frac{d}{dx}\left(\frac{dy}{dt}\right)+n\left(n-1\right){x}^{n-2}\frac{dy}{dt}\phantom{\rule{0ex}{0ex}}=n{x}^{n-1}\frac{d}{dt}\left(\frac{dy}{dt}\right)\frac{dt}{dx}+n\left(n-1\right){x}^{n-2}\frac{dy}{dt}\phantom{\rule{0ex}{0ex}}=n{x}^{n-1}\frac{{d}^{2}y}{d{t}^{2}}n{x}^{n-1}+n\left(n-1\right){x}^{n-2}\frac{dy}{dt}\phantom{\rule{0ex}{0ex}}={n}^{2}{x}^{2n-2}\frac{{d}^{2}y}{d{t}^{2}}+n\left(n-1\right){x}^{n-2}\frac{dy}{dt}$
$\therefore x\left({n}^{2}{x}^{2n-1}\frac{{d}^{2}y}{d{t}^{2}}+n\left(n-1\right){x}^{n-2}\frac{dy}{dt}\right)-n{x}^{n-1}\frac{dy}{dt}-4{x}^{3}y=0$
${n}^{2}{x}^{2n-1}\frac{{d}^{2}y}{d{t}^{2}}+n\left(n-1\right){x}^{n-1}\frac{dy}{dt}-n{x}^{n-1}\frac{dy}{dt}-4{x}^{3}y=0$
${n}^{2}{x}^{2n-1}\frac{{d}^{2}y}{d{t}^{2}}+n\left(n-2\right){x}^{n-1}\frac{dy}{dt}-4{x}^{3}y=0$
${n}^{2}{x}^{2n-4}\frac{{d}^{2}y}{d{t}^{2}}+n\left(n-2\right){x}^{n-4}\frac{dy}{dt}-4y=0$
${n}^{2}{t}^{\frac{2n-4}{n}}\frac{{d}^{2}y}{d{t}^{2}}+n\left(n-2\right){t}^{\frac{n-4}{n}}\frac{dy}{dt}-4y=0$
The ODE has the simplest form when we choose $n=2$
The ODE becomes
$4\frac{{d}^{2}y}{d{t}^{2}}-4y=0$
$\frac{{d}^{2}y}{d{t}^{2}}-y=0$
The auxiliary equation is
${\lambda }^{2}-1=0$
$\lambda =±1$
$\therefore y={C}_{1}{e}^{t}+{C}_{2}{e}^{-t}$