<s>[INST] Solve:$(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$ 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!
 [/INST] $$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$
$$yx'-x=y\sqrt{x^2 - y^2}$$
$$\left(\dfrac xy \right)'=\sqrt{\dfrac {x^2}{y^2} -1}$$
$$w'=\sqrt{w^2 -1}$$
Where $w=\dfrac xy$.Then integrate.
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