I have the differential equation: $(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$ where $p=\frac{dy}{dx}$
I have solved this up until $\frac{dy}{dx}$, but I am not able to reduce it to an exact differential equation.
Steps until here:
$$\frac{1}{x^2}\Bigl((x^2-x^2 y^2 +y^4)p^2- (2xy)p + y^2\Bigr)=0$$ As $\frac{1}{x^2}$ cannot equal zero, $$(x^2-x^2 y^2 +y^4)p^2- (2xy)p + y^2=0$$ Applying the quadratic formula: $$p=\frac{2xy \pm \sqrt{4x^2y^2-4x^2y^2+4x^2y^4-4y^6}}{2(x^2-x^2 y^2 +y^4)}$$ $$p=\frac{xy \pm y^2\sqrt{x^2-y^2}}{x^2-x^2 y^2 +y^4}$$ $$p=\frac{y\Bigl(x \pm y\sqrt{x^2-y^2}\Bigr)}{x^2-y^2(x^2-y^2)}$$ $$p=\frac{y\Bigl(x \pm y\sqrt{x^2-y^2}\Bigr)}{\Bigl(x-y\sqrt{x^2-y^2}\Bigr)\Bigl(x+y\sqrt{x^2-y^2}\Bigr)}$$ Thus, $$p=\frac{dy}{dx}=\frac{y}{\Bigl(x-y\sqrt{x^2-y^2}\Bigr)}$$ Or, $$p=\frac{dy}{dx}=\frac{y}{\Bigl(x+y\sqrt{x^2-y^2}\Bigr)}$$
From here, we can write the two equations as: $$xdy-ydx=y\sqrt{x^2 - y^2}dy$$ $$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$
I tried reducing this equation to an exact differential equation, but failed to do so. Kindly help me proceed from here. Thank you!
$$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$ $$yx'-x=y\sqrt{x^2 - y^2}$$ $$\left(\dfrac xy \right)'=\sqrt{\dfrac {x^2}{y^2} -1}$$ $$w'=\sqrt{w^2 -1}$$ Where $w=\dfrac xy$.Then integrate.