Finding a non constant solution to $ (x')^2+x^2=9 $ 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.
 A: Hint: try differentiating your equation with respect to $t$ - and use $x'\neq 0$

You have $(x')^2+x^2=9$
Differentiate with respect to $t$ to obtain $$2x'x''+2xx'=0$$
Divide through by $2x'$ to get $$x''+x=0$$
This is a second order linear differential equation, which should have a familiar solution. You then have to put the general solution back into the original equation to find a solution which works for that.
A: Hint: $$2x'(x''+x)=0.$$
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A: 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!
A: Brute force:
$$\begin{align}(x'(t))^2+(x(t))^2= 9&\implies x'(t)=\pm \sqrt{9-(x(t))^2}\\
&\implies \dfrac{x'(t)}{\sqrt{9-(x(t))^2}}=\pm 1\\
&\implies \dfrac{1}{3}\dfrac{x'(t)}{\sqrt{1-\left(\frac{x(t)}{3}\right)^2}}=\pm 1 \end{align}$$
Now use $\arcsin'(s)=\frac{1}{\large \sqrt{1-s^2}}$.
