How to solve the ODE: $ \frac{dy}{dx}=\frac{\sqrt{1-x^2}}{x} $

Question

I am learning differential equations and have been stuck on :

$$ \dfrac{dy}{dx}=\frac{\sqrt{1-x^2}}{x} $$

I tried using $\cos{t}, \sin{t}$ but failed.. how to solve this? please help.

Answer 1

Considering $$I=\int \dfrac{\sqrt{1-x^2}}{x}\,dx$$ let us try $x=\sin(t)$, $dx=\cos(t)dt$ which make $$I=\int \frac{\cos^2(t)}{\sin(t)}\,dt=\int \frac{1-\sin^2(t)}{\sin(t)}\,dt=\int \frac{dt}{\sin(t)}-\int \sin(t)\,dt=\int \frac{dt}{\sin(t)}+\cos(t)$$ So, now, considering $$J=\int \frac{dt}{\sin(t)}$$ using the tangent half-angle substitution $z=\tan(\frac t 2)$ makes $$J=\int\frac{dz}z=\log(z)=\log \left(\tan \left(\frac{t}{2}\right)\right)$$ Now, go back to $x$.

I am sure that you can take from here.