# Integral with square root of function of function

I have the function $y=y(x)$ with $y'=dy/dx$, and the following equation: $ky'=\pm\sqrt{k^{2}-y^{2}}$, where $k$ is constant.

Integrating this, given that $y(0)=0$, should give: $y=k\sin(x/k)$.

I don't know how such an integration was calculated and how we arrived at this result. Any help explaining the integration process would be appreciated.

Many thanks.

• Can we assume $k$ is a constant – Vishwa Iyer Jul 5 '14 at 16:41
• Yes, $k$ is constant. Sorry I forgot to mention that. – user135626 Jul 5 '14 at 16:41
• It is basically the fact that $\int \frac{dt}{\sqrt{1-t^2}}=\arcsin t+C$. – André Nicolas Jul 5 '14 at 16:42
• but how exactly do you handle this, since the exact form of $y(x)$ (i.e. how it depends on $x$) here is not known during integration? – user135626 Jul 5 '14 at 16:46

Assuming $k$ is a constant, you have $$k*\frac{dy}{dx} = \pm\sqrt{k^2-y^2}$$ $$\frac{k*dy}{\pm\sqrt{k^2-y^2}} = dx$$ Integrating both sides we get $$k\arcsin\left(\frac{y}{k}\right) = x +C$$ This is because $$\int \frac{du}{\sqrt{a^2-u^2}} = \arcsin(\frac{u}{a}) + C$$ For a constant $a$.
Then just solve for $y$.
• You dropped a $\pm$ when you integrated... :) (But +1 anyway) – apnorton Jul 5 '14 at 17:15