How to solve this differential equation of first order and higher degree solvable for y

Question

The question is $$x-yp=p^2$$. I arrived at the value of x which is $$x=\frac{-p^2}{2}+\frac{p}{2√(p^2-1)}log|p+√(p^2-1)|$$ But the answer is given as $$x=\frac{p}{√(p^2-1)}(c+ sin^-1p)$$ Where $p=\frac{dy}{dx}$

Answer 1

We have a d'Alembert's equation $y=xf(p)+g(p)$, where $f(p)=1/p$ and $g(p)=-p$. Differentiating with respect to $x$, $$ p=\frac{p-xp'}{p^2}-p' $$ and rearrange, $$ \frac{\mathrm{d}x}{\mathrm{d}p}=\frac1{p'}=\frac{x+p^2}{p(1-p^2)} $$ i.e., $$ \frac{\mathrm{d}x}{\mathrm{d}p}-\frac1{p(1-p^2)}x=\frac{p}{(1-p^2)}. $$ This is a first-order linear DE. With the integrating factor $\sqrt{1-p^2}/p$, we get $$ \frac{\mathrm{d}}{\mathrm{d}p}\frac{x\sqrt{1-p^2}}{p}=\frac1{\sqrt{1-p^2}} $$ giving the answer $$ x=\frac{p}{\sqrt{1-p^2}}(c+\sin^{-1}p). $$