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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.

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    One question mark is enough.2010-11-23
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    Mariano, Mike, sorry by question mark.2010-11-23

3 Answers 3

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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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    It sounds great, but you solution is not enough general.2012-06-16
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    @doraemonpaul What do you mean by "the solution is not general enough"?2012-06-16
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$\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 $\rm\ y'\ =\ \pm 2\:x\ y\ \ \Rightarrow\ \ y\ =\ c\ e^{\pm x^2}$

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    This is amazing, thank you2010-11-23
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    How does it follow that $f = \pm 2x$? It is one of the solutions, but couldn't there be others? For instance, even though it is trivial, $y=0$ admits _any_ $f$.2010-11-23
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    @Moron: Hint: uniqueness theorem2010-11-23
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    @Bill: What is the motivation for $y'=fy$ ?2010-11-23
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    seeing the solution? ;) Seriously though it is equivalent to guessing $y = e^{f(x)}$ and then finding $f'(x)$, which might not occur to one without knowing in advance.2010-11-23
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    @Bill: I still don't see it. Perhaps you would care to update your answer? Thanks.2010-11-23
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    It sounds great, but you solution is not enough general.2012-06-16
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    @doraemonpaul Your misread the notation. As should be clear from the derivation, the constant is not meant to be the same for both solutions. It denotes the same solution space as in your recent answer.2012-06-16
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$y''-\dfrac{y'}{x}=4x^2y$

$x\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}-4x^3y=0$

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

$\lambda^2-1=0$

$\lambda=+-1$

$\therefore y=C_1e^t+C_2e^{-t}$

$y=C_1e^{x^2}+C_2e^{-x^2}$