Can one solve this differential equation without prior knowledge?

Question

Simple question; how can one prove that the solution to $y'^2 = 1 - y^2$ is $y = \pm\sin(x+c), \ c \in \Bbb{R}$, without prior knowledge of the solution?

Answer 1

$$(y')^2=1-y^2\Rightarrow \Big(\frac{dy}{dx}\Big)^2=1-y^2\Rightarrow |\frac{dy}{dx}|=\sqrt{1-y^2}$$ So $\frac{dy}{dx}=\sqrt{1-y^2}$ or $\frac{dy}{dx}=-\sqrt{1-y^2}$ Now we can solve this by direct integration $$\frac{dy}{dx}=\sqrt{1-y^2}\Rightarrow \frac{dy}{\sqrt{1-y^2}}=dx\Rightarrow \int\frac{dy}{\sqrt{1-y^2}}=\int dx\Rightarrow \arcsin(y)=x+c$$ therefore $$y=\sin(x+c)$$ the other solution would be $y=\sin(-x+c)$.

Answer 2

It's a nice exercise in complex analysis that if $f(x)^2 + y(x)^2 = 1$, where $f$ and $y$ are analytic functions in some domain, then $f(x) = \sin(h(x))$ and $y(z) = \cos(h(x))$ for some analytic function $h$ in that domain. Now if $y = \sin(h(x))$, $y' = h'(x) \cos(h(x))$, so if that is $\pm \cos(h(x))$ we have $h'(x) = \pm 1$. Thus for some constant $c$ we have $h(x) = \pm (x + c)$, and $y = \sin(\pm( x+c)) = \pm \sin(x+c)$.

Note also that you can get rid of the $\pm$ because $\sin(x+\pi) = -\sin(x)$.