Problems solveing an ODE

Question

I'm having problems solving the following nonlinear second-order differential equation: $$y''=Ay\left(1+y'^2\right)^{3/2} + \left(1+y'^2\right)^{2}$$ I've tried the traditional methods when the independent variable is missing but couldn't solve it anyways. Every help is welcome.

Answer 1

Assume $A\neq0$ for the key cases:

Let $u=y'$ ,

Then $u\dfrac{du}{dy}=Ay(1+u^2)^\frac{3}{2}+(1+u^2)^2$

Let $v=1+u^2$ ,

Then $\dfrac{dv}{dy}=2u\dfrac{du}{dy}$

$\therefore\dfrac{1}{2}\dfrac{dv}{dy}=Ayv^\frac{3}{2}+v^2$

$\left(y+\dfrac{\sqrt v}{A}\right)\dfrac{dy}{dv}=\dfrac{1}{2Av^\frac{3}{2}}$

This belongs to an Abel equation of the second kind.

Let $w=y+\dfrac{\sqrt v}{A}$ ,

Then $y=w-\dfrac{\sqrt v}{A}$

$\dfrac{dy}{dv}=\dfrac{dw}{dv}-\dfrac{1}{2A\sqrt v}$

$\therefore w\left(\dfrac{dw}{dv}-\dfrac{1}{2A\sqrt v}\right)=\dfrac{1}{2Av^\frac{3}{2}}$

$w\dfrac{dw}{dv}-\dfrac{w}{2A\sqrt v}=\dfrac{1}{2Av^\frac{3}{2}}$

$w\dfrac{dw}{dv}=\dfrac{w}{2A\sqrt v}+\dfrac{1}{2Av^\frac{3}{2}}$

Let $x=\dfrac{\sqrt v}{A}$ ,

Then $\dfrac{dw}{dv}=\dfrac{dw}{dx}\dfrac{dx}{dv}=\dfrac{1}{2A\sqrt v}\dfrac{dw}{dx}$

$\therefore\dfrac{w}{2A\sqrt v}\dfrac{dw}{dx}=\dfrac{w}{2A\sqrt v}+\dfrac{1}{2Av^\frac{3}{2}}$

$w\dfrac{dw}{dx}=w+\dfrac{1}{v}$

$w\dfrac{dw}{dx}=w+\dfrac{1}{A^2x^2}$

This belongs to an Abel equation of the second kind in the canonical form.