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!


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

  • $\begingroup$ so I managed to get that $q=\tan{(x+c)}$, and plug it in your equation to give me $p=x*\tan{(x+c)}$. I'm not sure whether I need to find the derivative of $p$ and then plug it into the original equation (because if I do I would only be left with a non-differential equation consisting of the variable $x$. Maybe I just need to plug the value $p$ into the original equation and then solve for $p'$? $\endgroup$ – john doe May 17 at 13:47
  • $\begingroup$ Sorry, I made a haphazard mistake in calculation and also failed to realize that I just need to plug the solution in $p=xq$. $\endgroup$ – john doe May 17 at 14:13

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

  • $\begingroup$ Why does $$ \dfrac {xp' -p}{x^2} = \left(\dfrac {p}{x} \right)'$$? It looks to me like you dropped a (p/x²) term there. $\endgroup$ – Brilliand May 17 at 22:43
  • $\begingroup$ Try to differentiate $f/g$ here $f=p$ and $g=x$ @Brilliand $\endgroup$ – Aryadeva May 17 at 22:48
  • $\begingroup$ It's the rule. When you differentiate a quotient of functions such as $f/g$ you get : $$\dfrac {f'g-fg'}{g^2}$$ @Brilliand Take a look here for the qotient rule web.iit.edu/sites/web/files/departments/academic-affairs/… $\endgroup$ – Aryadeva May 17 at 22:54
  • $\begingroup$ Oh, got it - I was wrong to treat x as a constant. $\endgroup$ – Brilliand May 17 at 22:57
  • $\begingroup$ No problem you're welcome @Brilliand $\endgroup$ – Aryadeva May 17 at 22:58

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