Solving the Equation (y^2)+(y')^2=1: Ideas and Guidance

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Homework Statement



Find a function of y whose square plus the square of its derivative is 1.
i.e. (y^2)+(y')^2=1 and carry out your ideas.


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The Attempt at a Solution



Can anyone just help me out with this one. Kinda confused by the questioning. Just point me in the right direction.
 
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Think trig identities...
 
I still don't see it. How does the derivative factor into it?
 
Dr. Lady is suggesting you simply guess the answer.
Otherwise you have y'=+/-sqrt(1-y^2). You can try to integrate that.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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