Solving a Riccati equation - doesn't fit into any category I tried until now

Question

I have this ODE:

$$p = xp' - x(p^2 + x^2)$$

after dividing by $x \neq 0$ we get

$$ p' = \frac{p}{x} + p^2 + x^2 $$

Which I recognized as a Riccati equation, and also confirmed this on WolframAlpha.

However, here's the problem. We always dealt with three subtypes of the Riccati equation:

a) For any Riccati equation $y' = P(x)y^2 + Q(x)y + R(x)$, if P(x), Q(x) and R(x) are constants, then it is a separable equation

b) If $y' = Ay^2 + \frac{B}{x}y + \frac{C}{x^2}$, A,B,C are constants, we let $z=yx, z=z(x)$

c) If a particular solution $y_1$ is known, we let $y(x) = y_1(x) + \frac{1}{z(x)}$

However, my equation does not fall into any of the three categories I mentioned and I don't know how to proceed. Any help would be appreciated!

Answer 1

Start with the change of variables $p=x q$, giving $$q+xq'=q+x^2(1+q^2). $$ Can you take it from here?

Answer 2

$$p = xp' - x(p^2 + x^2)$$ $$xp' -p = x(p^2 + x^2)$$ $$ \dfrac {xp' -p}{x^2} = x\left(\dfrac {p^2}{x^2} + 1\right)$$ $$\left(\dfrac {p}{x} \right)' = x\left(\dfrac {p^2}{x^2} + 1\right)$$ This is a separable DE. $$\arctan \left (\dfrac px \right)=\dfrac {x^2}{2}+C$$ $$p=x\tan \left (\dfrac {x^2}{2}+C\right)$$