Differential equation of the form: $y'' - \frac{1}{(1+x^2)^2} y=-\frac{Ax}{\sqrt{1+x^2}}$

Question

When attempting this problem on physics SE, I came across this differential equation: $$ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{y}{\left( 1+x^2 \right)^2}+A\frac{x}{\sqrt{1+x^2}} = 0 $$ where the constant $A$ is greater than $0$. Wolfram Alpha says the solution is $$ \left( -\frac{1}{2}Ax + c_1 + c_2\tan^{-1}x \right)\sqrt{1+x^2} $$ but I am not able to derive the result myself.

I have tried substituting $x=\tan u$, so that the equation becomes $$ \frac{\mathrm{d}^2y}{\mathrm{d}u^2} - y = -A\frac{\sin u}{\cos^{4}u} $$ but then variation of parameters by setting $y=y_1 \eta + y_2\zeta$ gives the monstrous integrals $$ \eta = -\frac{A}{2B}\int \frac{e^{-u}\sin u}{\cos^{4}u}\mathrm{d}u, \quad \zeta = -\frac{A}{2B}\int \frac{e^{u}\sin u}{\cos^{4}u}\mathrm{d}u $$ which Wolfram Alpha says can only be evaluated using ellptic integrals.

Answer 1

For the general equation

$$ y'' - \frac{y}{(1+x^2)^2} = 0 $$

I'm not sure of a general technique which applies, but it's reasonable to guess $y = (1+x^2)^C$ is a solution.

$$ y' = 2Cx (1+x^2)^{C-1} $$ $$ y'' = 2C (1+x^2)^{C-1} + 4C(C-1) x^2(1+x^2)^{C-2} $$ $$ \frac{y''}{y} = \frac{1}{(1+x^2)^2} = \frac{2C}{1+x^2} + \frac{4C(C-1)x^2}{(1+x^2)^2} = \frac{(4C^2-2C)x^2 + 2C}{(1+x^2)^2} $$

This is satisfied for every $x$ if $C=\frac 12$, so $y_1(x) = \sqrt{1+x^2}$ is one solution. The other independent general solution can be found by taking $y_2(x) = v(x) y_1(x)$, but moving on to the particular case will also produce that term.

So to solve the original

$$ y'' - \frac{y}{(1+x^2)^2} + A\frac{x}{\sqrt{1+x^2}} = 0 $$

suppose $y(x) = v(x) (1+x^2)^{1/2}$.

$$ y' = v' (1+x^2)^{1/2} + xv (1+x^2)^{-1/2} $$

$$ y'' = v'' (1+x^2)^{1/2} + 2xv' (1+x^2)^{-1/2} + v (1+x^2)^{-1/2} - x^2 v (1+x^2)^{-3/2} $$ $$ y'' = v'' (1+x^2)^{1/2} + 2xv' (1+x^2)^{-1/2} + v (1+x^2)^{-3/2} $$ $$ y'' - y(1+x^2)^{-2} + Ax(1+x^2)^{-1/2} = 0 $$ $$ v''(1+x^2)^{1/2} + (2v'+A)x (1+x^2)^{-1/2} = 0 $$ $$ \frac{v''}{2v'+A} = -\frac{x}{1+x^2} $$ $$ \frac 12 \ln(2v'+A) = -\frac 12 \ln(1+x^2) + C_1 $$ $$ 2v'+A = \frac{e^{2C_1}}{1+x^2} $$ $$ 2v + Ax = e^{2C_1} \tan^{-1}x + C_2 $$ $$ v = \frac{e^{2C_1}}{2} \tan^{-1} x + \frac{C_2}{2} - \frac A2 x $$

And finally,

$$ y(x) = v \sqrt{1+x^2} = \sqrt{1+x^2} \left(c_1 \tan^{-1} x + c_2 - \frac A2 x\right) $$

Answer 2

I have tried substituting $x=\tan u$, so that the equation becomes $$ \frac{\mathrm{d}^2y}{\mathrm{d}u^2} - y = -A\frac{\sin u}{\cos^{4}u} $$

This is incorrect. After you do the sub $x=\tan u$, you should get

$$\frac{d^2y}{du^2}\color{red}{-2\tan (u) \frac{dy}{du}} - y = -A\frac{\sin u}{\cos^{4}u}$$

where you miss the first order derivative term which is colored in red.