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.
Best Answer
Yes go ahead $$y\sqrt{1+p^2}=ap\sqrt{1+p^2}-a\sinh^{-1}{p}+C \implies y(p)=ap+\frac{C-\sinh^{-1}p}{\sqrt{1+p^2}}.$$ put this in $$x(p)=ap^2-py(p)$$ to get $x(p)$.
Finally $x(p)$ and $y(p)$ constitue the parametric solution of the ODE, where $p$ acts as a parameter only.$C$ is the integration-constant,
Note that $$\int \frac{p^2}{\sqrt{1+p^2}}dp= \int \left(\sqrt{1+p^2}-\frac{1}{\sqrt{1+p^2}}\right)dp=\frac{1}{2}\left( p\sqrt{1+p^2}-\sinh^{-1} p\right)$$ Both the integrals are standard ones.