Solution of :$y''=y'(1+y^2)$

Question

I have tried to solve this differential equation $y''=y'(1+y^2)$ as shown in my below attempt in the attached paper, but I didn't succeed , I want at a least how I can complete solution from the last step , or to use other methods and thanks ?

Answer 1

This is a cheap but useful trick:

$$y''=\frac{dy'}{dx}=\frac{dy'}{dy}\frac{dy}{dx}=y'\frac{dy'}{dy}$$

Your equation becomes:

$$y'\frac{dy'}{dy}=y'(1+y^2)$$

$$\frac{dy'}{dy}=(1+y^2)$$

$$dy'=(1+y^2)dy$$

Integrating this is simple:

$$y'=y+\frac{y^3}{3}+C$$

You won't get much further if $C$ is not zero.

But if $C$ is zero, and that is possible for fine tuned values of initial conditions $y(0)$ and $y'(0)$, you will be able to proceed:

$$ \frac{dy}{dx}=y+\frac{y^3}{3}$$

$$x=\int \frac{dy}{y+\frac{y^3}{3}}$$

This integral is not a rocket science and can be solved pretty easily:

$$x=\ln (y)-\frac{1}{2} \ln \left(y^2+3\right)+\ln{C}$$

$$x=\ln{\frac{Cy}{\sqrt{y^2+3}}}$$

$$e^x = \frac{Cy}{\sqrt{y^2+3}}$$

$$e^{2x} = \frac{C^2y^2}{y^2+3}$$

Use initial conditions to evaluate $C$. Solve the last equation with respect to $y$ (which is easily doable) and you are done.

If the initial conditions are not fine tuned, you are doomed.

Answer 2

You can also directly integrate to the equation you found, as the time derivative of $y+\frac13y^3$ is $(1+y^2)y'$, so that $$ y'=\frac13y^3+y+C $$ directly follows by time integration of the original equation, no fancy substitutions needed.

Indeed for $C=0$ you have a Bernoulli-equation which is a linear equation in $v=y^{-2}$ while for $C\ne 0$ one can proceed via partial fraction decomposition.