TITLE: Finding a non constant solution to $ (x')^2+x^2=9 $
QUESTION [6 upvotes]: How do I find a non-constant solution this equation? I've tried to solve for $x$, but the final answer should be in the form of $x(t)=...$
$ (x')^2+x^2=9 $
I'm not sure where to start.

REPLY [14 votes]: How about this:  notice that $(x')^2 + x^2 = 9$ is a circle in $x$-$x'$ space, a circle of radius $3$.
Then the point $(\frac{x}{3}, \frac{x'}{3})$ lies on the unit circle.  That means there will be some  function of $t$, $\theta(t)$, such that $x(t) = 3\cos \theta (t)$ and $x'(t) = 3\sin \theta (t)$.   Based on this information, and assuming $\theta(t)$ is differentiable (which can be proved via the implicit function theorem), we have $x'(t) = -3\theta'(t) \sin \theta (t)$ as well from which we infer $\theta'(t) = -1$, leading to $\theta(t) = c - t$ for some constant $c$.  So $x(t) = 3\cos(c - t)$.
Hope this helps.  Cheers!