# 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.

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

• Do you mean integrate both sides using that identity? – Bob Shannon Sep 5 '13 at 19:50
• @Bob That is what I mean. – Git Gud Sep 5 '13 at 19:52
• Still not sure what's going on. I've integrated both sides and got arcsin(x(t)/3) = x(t). – Bob Shannon Sep 5 '13 at 20:02
• Your RHS is wrong. Can you fix it and follow up? – Git Gud Sep 5 '13 at 20:04
• arcsin(x(t)/3) = t? – Bob Shannon Sep 5 '13 at 20:05

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!

• Purdy.${{{{}}}}$ – Git Gud Sep 5 '13 at 20:02
• @Git Gud: Thank you, thank you! – Robert Lewis Sep 5 '13 at 20:04
• @nbubis: glad you enjoyed it. I exploited similar technique in math.stackexchange.com/questions/473458/…, but in a much more sophisticated way. Shameless self-promotion! ;) – Robert Lewis Sep 6 '13 at 1:54

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.

• There is no t variable to integrate, which is what is throwing me off. – Bob Shannon Sep 5 '13 at 20:03
• @bob - differentiate, don't integrate! – Mark Bennet Sep 5 '13 at 20:07
• Note that $x'=0$ is also a possible solution, telling us there aren't any more solutions than sines, cosines, and constant functions. – abnry Sep 5 '13 at 20:32
• @nayrb The question asks for a non-constant solution, my explanation uses the hint, which includes $x'\neq 0$ – Mark Bennet Sep 5 '13 at 20:34
• Yup, just commenting that sines and cosines are the only non-constant solutions. – abnry Sep 5 '13 at 20:35

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

• What do I do from here? How did you figure that out? – Bob Shannon Sep 5 '13 at 20:02
• The general solution of $x''(t)+x(t)=0$ is $C_1\cos t + C_2 \sin t$. – njguliyev Sep 5 '13 at 20:33