Find a function whose square plus the square of its derivative is 1.

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liingliing8 · Accepted Answer

We want to find a function y(x) that satisfy the equation
$y^{2} + {(\frac{d y}{d x})}^{2} = 1 \to {(\frac{d y}{d x})}^{2} = 1 - y^{2}$
$∴ \frac{d y}{d x} = \pm \sqrt{1 - y^{2}}$
Which is a seperable DE, thus we multiply both sides by $\frac{d y}{\sqrt{1 - y^{2}}}$ to get
$\frac{d y}{\sqrt{1 - y^{2}}} = d x \to \int \frac{d y}{\sqrt{1 - y^{2}}} = \int d x$
From appendix I in textbook we find that
$\int \frac{d u}{\sqrt{1 - u^{2}}} = \sin^{- 1} u + C$
Applying this formula we get
$\sin^{- 1} y = x + C \to y = \sin (x + C)$
Thus, any solution of the form $y = \sin (x + C)$ satisfy the required condition, from the DE, we note that $y = 1$ and $y = - 1$ are singular solutions which make the denominator $\sqrt{1 - y^{2}} = 0$
If the negative sign was used we get
$y = \sin (C - x) \to y = - \sin (x + C)$
Result: $y = \pm \sin (x + C)$

Find a function whose square plus the square of its derivative is 1.

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