Magic of 4

Consider the solution of the equation to be $y = uv$ .

And the differential equation to be of the form:-

$y'' + Py' +Qy = F$ [ Where P,Q and F are functions of x]

Now evaluate $y'$ and $y''$ in terms of $u',u'',v',v'',u,v$

[Here f' represents $\frac{df}{dx}$ and f'' means $\frac{d^{2}f}{dx^{2}}$ respectively]

So the equation transforms to

$\displaystyle uv'' + u(P+ \frac{2}{u}u')v' +(u'' +Pu' +Qu)v = F$

Choose u such that the coefficient of $v'$ becomes $0$ .

So we have $P+ \frac{2}{u}u' = 0$

So This is a first order linear differential equation and can be easily solved

The solution is $\large u = e^{\frac{-1}{2} \int P dx}$

Now substitute this in the equation and we have :-

$\displaystyle u'' = \frac{-1}{2} (Pu' + uP')$

$\displaystyle u'' = \frac{P^{2}}{4}u - \frac{u}{2}P'$

So the final equation becomes :-

$v'' + Iv = S$

Where $I = Q-\frac{1}{4}P^{2} - \frac{1}{2}P'$

And $\large S=F e^{\frac{1}{2} \int P dx}$

This is a general method and can be applied in this equation .

Here $P = \frac{1}{x}$ and $Q= 4-\frac{1}{4x^{2}}$

So here $\large u=e^{\int \frac{-1}{2x}dx}$

So $u=\frac{1}{\sqrt{x}}$

So now solving for v we have

$v'' + Iv = 0$

$I = 4-\frac{1}{4x^{2}} - \frac{1}{4x^{2}} + \frac{1}{2x^{2}}$ [using the formula we derived earlier]

So $I = 4$

So the equation for $v$ just becomes

$v'' + 4v =0$

Which is a differential equation with constant coefficient and can be solved easily using the known methods to evaluate the Complementary Functions

The roots of the Auxiliary equation is $2i$ and $-2i$

So the complementary function is just $v=C\cos(2x) + K\sin(2x)$ Where $C$ and $K$ are arbitrary constants

And the particular integral for this is just $0$ because the RHS is $0$

So the complete solution of the differential equation is

$y= \frac{1}{\sqrt{x}}(C\cos(2x) + K\sin(2x))$

Now just using the information given in the question we see that $K=C=\pi\sqrt{\pi}$

So our answer becomes

$\large \frac{\pi\sqrt{\pi}}{\sqrt{10}} (\cos(20) + \sin(20))$

Here the $"20"$ inside the $\sin$ and $\cos$ are taken in radians