I have looked all over the place and it seems I can't find a reference for solution methods to common forms of first order nonlinear ordinary differential equations (ODEs). Currently, I am starting at the following population model

$$ \dot P = a P^2 - b P $$

given $a, b > 0$, and not really getting anywhere with this. It seems that Wolfram|Alpha cannot parse it so I can't search the solution steps for insight, though maybe there's some special syntax I should be using there.

How would I go about solving this one in particular, and where would I find a compendium of ODE solution methods?

  • $\begingroup$ Did the answer below resolve your issues? If so, you should upvote and/or accept the answer. Regards $\endgroup$
    – Amzoti
    Oct 3, 2013 at 16:10

1 Answer 1



  • The equation is separable.
  • Write:

$$\displaystyle \int \frac{dP}{a P^2 - b P} = \int dt$$

Can you take it from here?

Book References

  • Nonlinear Ordinary Differential Equations, Jordan and Smith

You might also want to visit Equation World.

  • $\begingroup$ Sure, so $ ln(aP^2 - bP) = t + C_1 $, then $ aP^2 - bP = Ce^t $ which is great except that I still have no idea what P(t) is, and I don't have an initial condition to get rid of the integration constant placeholder C. Even if I'm being particularly clever and noting that $ aP^2 - bP = Ce^t $ could be substituted into the initial differential equation, I'm left with $ dP/dt = Ce^t $ which becomes $ P = C_1 e^t + C_2 $ after integration, so I'm even worse off than before. Where am I going wrong here? $\endgroup$
    – user98131
    Oct 2, 2013 at 2:22
  • $\begingroup$ Did you do partial fractions on the LHS? The final result should be $P(t) = \dfrac{b}{a + e^{b(t + c)}}$. Try substituting this result back into the original. What do you get? Regards $\endgroup$
    – Amzoti
    Oct 2, 2013 at 2:37
  • $\begingroup$ Nice work, @Amzoti! +1 $\endgroup$
    – amWhy
    Oct 2, 2013 at 11:28
  • $\begingroup$ I'll be thinking about you! Hope it's nothing serious! My visits are going fine: same ol' same ol'! But I'm striking out on the little time I've had for MSE! $\endgroup$
    – amWhy
    Oct 2, 2013 at 12:49

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