Solve $\operatorname{x}\operatorname{dy}-\operatorname{y}\operatorname{dx}=\operatorname{x}\sqrt{x^2-\operatorname{y^2}}\operatorname{dy} $

Question

How do I show that $$\operatorname{x}\operatorname{dy}-\operatorname{y}\operatorname{dx}=\operatorname{x}\sqrt{x^2-\operatorname{y^2}}\operatorname{dy} $$

has a primitive of:

$$ \operatorname{y=x}\sin(y+C) $$

I must be missing something pretty clear because I'm pretty sure this isn't supposed to be a difficult problem.

I've tried finding an integrating factor to change it into an exact equation, but nothing is popping out at me, nor is it homogeneous. I can't seem to write it as a Bernoulli Equation, and can't seem to find a way. I'm sure I'm just missing a glaring problem considering how simple the solution is. Thanks in advance!

Answer 1

$$\operatorname{x}y'-\operatorname{y}=\operatorname{x}\sqrt{x^2-\operatorname{y^2}}y'$$ $$(\frac yx)'=y'\sqrt{1-(\frac yx)^2}$$ $$\int \frac {d(\frac yx)}{\sqrt{1-(\frac yx)^2}}=\int y'dx $$ It's easy to integrate now $$\int \frac {d(\frac yx)}{\sqrt{1-(\frac yx)^2}}=y+K $$

$$\arcsin(\frac yx)=y+K $$ $$y=x\sin(y+K) $$ Are you sure that the primitive you have is correct ?

Answer 2

Let $y=vx\implies dy=v+xdv$

So, for $y\ge0,$ we have $$vydy-y(ydv+vdy)=vy|y|\sqrt{v^2-1}dy$$

$$\implies\dfrac{dv}{v\sqrt{v^2-1}}=-dy\cdot\text{ sign}(y)$$

Integrating both sides we get $$K-y=\text{arcsec}(v)=\arccos\dfrac1v=\dfrac\pi2-\arcsin\dfrac1v$$