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How do I solve the differential equation $r(t)^2 + r^{'}(t)^2 = 1$, where $r: \mathbb R \rightarrow \mathbb R$ is a smooth real-valued function ?

In Calculus I've seen linear (higher-order) differential equations, but never equations where the functions involved are squared.

However, in some other mathematical textbook, I'm reading, the differential equation above is to be found.

Can I use my knowledge from Calculus to solve differential equation completely for all solutions ? Please show me how or describe the process.

Shuzheng
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    Note that $dr/dt = \pm\sqrt{1-r^2}$. Can you take it from there? – Simon S Jan 29 '15 at 17:35
  • Nearly a duplicate: http://math.stackexchange.com/questions/485115/finding-a-non-constant-solution-to-x2x2-9 – Git Gud Jan 30 '15 at 00:07
  • This question is related to this earlier question by the OP. I provided this sketchy solution in a comment: $(dr/dt)^2+r^2=1\implies dr/dt=\sqrt{1-r^2}\implies$ $dr/\sqrt{1-r^2}=dt\implies\int dr/\sqrt{1-r^2}=\int dt\implies$ $\arcsin(r)=t+C\implies r=\sin(t+C)$. I take it that the OP wants more details. – Joonas Ilmavirta Jan 30 '15 at 09:00
  • How do I argue that $\sin(t+C)$ is the unique solution ? I've tried considering $(\frac {y} {\sin(t+C)})'$ is $y$ is another solution etc. – Shuzheng Jan 30 '15 at 09:02
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    There are uniqueness and existence theorems for ODEs: example pdf. I would suggest trying to use that general machinery rather than doing something more ad hoc. – Joonas Ilmavirta Jan 30 '15 at 09:04

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