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.

  • $\begingroup$ Can we assume $k$ is a constant $\endgroup$ – Vishwa Iyer Jul 5 '14 at 16:41
  • $\begingroup$ Yes, $k$ is constant. Sorry I forgot to mention that. $\endgroup$ – user135626 Jul 5 '14 at 16:41
  • $\begingroup$ It is basically the fact that $\int \frac{dt}{\sqrt{1-t^2}}=\arcsin t+C$. $\endgroup$ – André Nicolas Jul 5 '14 at 16:42
  • $\begingroup$ 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? $\endgroup$ – 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 $y$.

  • $\begingroup$ You dropped a $\pm$ when you integrated... :) (But +1 anyway) $\endgroup$ – apnorton Jul 5 '14 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.