Second order nonlinear differential equation $x''+Hx =A(1-J/(2x^2))$

Question

I have arrived at a differential equation and I need to solve for $x$. $$ \frac{\mathrm{d}^2x}{\mathrm{d}E^2}+Hx =A\left(1-\frac{J}{2x^2}\right) $$ where $H$, $A$, and $J$ are constants.

I know that I can use elliptic integrals, but I need some help with the integration steps from here on.

Answer 1

Hint

Solve the equation with the RHS equal to zero and then use the variation of parameters method.

https://en.m.wikipedia.org/wiki/Variation_of_parameters

Answer 2

$d^2x \over dE^2$+$Hx$ =$a$($1$+$J\over x^4$ -$1 \over {2x^2}$)

$\frac{dx}{dE}=\frac{1}{\frac{dE}{dx}}$

$\frac{d^2 x}{dE^2}=\frac{d}{dx}\left(\frac{dx}{dE}\right)\frac{dx}{dE} = -\frac{\frac{d^2E}{dx^2}}{\left(\frac{dE}{dx}\right)^2}\frac{dx}{dE} = -\frac{\frac{d^2E}{dx^2}}{\left(\frac{dE}{dx}\right)^3}$

$-\frac{\frac{d^2E}{dx^2}}{\left(\frac{dE}{dx}\right)^3}=-Hx+a\left(1+\frac{J}{x^4}-\frac{1}{2x^2}\right)$

After integration : $\frac{1}{2\left(\frac{dE}{dx}\right)^2}=-\frac{H}{2}x^2+ax-a\frac{J}{3x^3}+a\frac{1}{2x}$+constant.

$\frac{dE}{dx}=\frac{1}{\sqrt{-Hx^2+ax-a\frac{2J}{3x^3}+a\frac{1}{x}+c_1}}$

$$E(x)=\int \frac{dx}{\sqrt{-Hx^2+ax-a\frac{2J}{3x^3}+a\frac{1}{x}+c_1}}$$ There is no closed form for this integral. So, further calculus must be numerical.