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.

Solution 1:

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!

Solution 2:

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 $\arcsin'(s)=\frac{1}{\large \sqrt{1-s^2}}$.

Solution 3:

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.

Solution 4:

Hint: $$2x'(x''+x)=0.$$ $ $ $ $