<s>[INST] Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$ Solve $x+yp=ap^2$ where $p=\frac {dy}{dx}$
My Attmept:
$$x+yp=ap^2$$
$$x=ap^2-yp$$
This is solvable for $x$ so differentiating both sides w.r.t $y$
$$\frac {dx}{dy} = 2ap\cdot \frac {dp}{dy} - y\cdot \frac {dp}{dy} -p$$
$$\frac {1}{p} + p = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {1+p^2}{p} = (2ap-y) \cdot \frac {dp}{dy}$$
$$\frac {dy}{dp} = \frac {2ap^2}{1+p^2} - \frac {yp}{1+p^2}$$
$$\frac {dy}{dp} + \frac {p}{1+p^2} \cdot y = \frac {2ap^2}{1+p^2}$$
$$\textrm {This is Linear so}$$
$$\textrm {Integrating Factor}=e^{\int \frac {p}{1+p^2} dp}=\sqrt {1+p^2}$$
So we have
$$y\sqrt {1+p^2} = \int \frac {2ap^2}{1+p^2} \times \sqrt {1+p^2} dp + c$$
$$y\sqrt {1+p^2} = 2a \int \frac {p^2}{\sqrt {1+p^2}} dp + c $$
I could not solve further from here.
 [/INST] Usually you get fastest to the relevant equations if you start with the derivative by $p$, using $\dot y(p)=p\dot x(p)$. Then
$$
\dot x+p\dot y+y=2ap
$$
Multiply with $p$ and eliminate $x$
$$
(1+p^2)\dot y+py=2ap^2\implies \sqrt{1+p^2}y(p)=\int \frac{2ap^2}{\sqrt{1+p^2}}dp
$$
Now substitute $p=\sinh(u)$ in the last integral,
$$
=\int a(\cosh(2u)-1)du
=a\sinh(u)\cosh(u)-au+C
=ap\sqrt{1+p^2}-a\sinh^{-1}(p)+C.
$$
</s>