Is there a solution to the differential equation $y'' = -y^3$?
For lesser powers there are obvious solutions, for example…
$y'' = -y\ \ \ \to \ \ \ y=\sin x$
$y'' = y \ \ \ \to \ \ \ y=e^x$
$y'' = -y^2\ \ \ \to \ \ \ y=\frac1x$
I don't think there's an easy way to solve my question though, or at least none that I can think of, because separate and integrate won't work, and the equation isn't linear so no characteristic equation fanciness will happen.
I just wanted to throw this out, and if a solution exists, or it is proven it can't exist, I'd greatly appreciate it!
BTW, I already found one solution. I just want a way to solve it without blatant assumptions. I assumed the solution would be of the form $y=a\cdot x^r$. Solving for variables in the equation led to the solution $y=\frac{\sqrt2}x$, but it's not a truly proper way to solve a differential equation. I just need a bit of help, or a push, or a shove if needed. Thanks!
Best Answer
if $y''$ is only a functionof $y$, the easiest "trick" is to multiply throughout by $y'$
$$ y'y'' = -y^3y' $$
then integrate to get
$$ \frac{y'^2}{2} = C - \frac{y^4}{4} $$
Solving for $y'$
$$ y' = \pm \sqrt{\frac{c-y^4}{2}}$$
which is separable. The solution does not have a closed form.