Consider the ODE $$\frac{dx}{dt} = ax + b$$ where $a$ and $b$ are two parameters. The way to solve this is to divide both sides by $ax+b$ and integrate:

$$\int \frac{\dot x}{ax+b}dt = t+C \\ \frac{\log|ax+b|}{a} = t+C \\ x(t) = Ke^{at}-\frac ba$$

Easy enough. But I'm not sure why we're not excluding some possible solutions in the first step of this approach. Doesn't dividing by $ax+b$ immediately rule out any solution where $x(t)=-\frac ba$ anywhere in the interval over which the function is defined? That seems like we might be losing a lot of potential solutions. So why is the above solution the general solution?

  • $\begingroup$ The constant solution $x(t)=-b/a$ is an equilibrium solution, so it would be handled separately. $\endgroup$ – Nicholas Stull Apr 22 '16 at 19:01
  • $\begingroup$ What about any function $x(t)$ to could potentially equal $-b/a$ at some finite number of points? Or even infinitely many points that just isn't all of them in the interval? $\endgroup$ – user333824 Apr 22 '16 at 19:02
  • $\begingroup$ "The way to solve this is" not to use separation of variables unnecessarily, but instead proceed in this fashion. But if you really want to use separation of variables, you can justify that those are the only solutions by using the Picard–Lindelöf theorem. $\endgroup$ – Git Gud Apr 22 '16 at 19:04
  • $\begingroup$ I see. So this is only the way my book solves it -- not the most rigorous way. Thanks @GitGud! $\endgroup$ – user333824 Apr 22 '16 at 19:06
  • $\begingroup$ @user333824 You're welcome. You might also be interested in this question. $\endgroup$ – Git Gud Apr 22 '16 at 19:17

You're right, and this is a good issue to point out. In this case, it's straightforward to show uniqueness, though: Suppose that $x(t)$ is a solution and notice that if $y(t) = e^{-at} x(t)$, we have

\begin{align*} y'(t) &= -ae^{-at} x(t) + e^{-at} x'(t) \\ &= -ae^{-at} x(t) + e^{-at} \big(ax(t) + b\big) \\ &= be^{-at} \end{align*}

Now integrating shows what $y$ must be, and hence $x$. No division by zero at all.

Sometimes, one can solve an equation in a somewhat ad hoc manner, as this separation of variables does, and then simply check that the solution is valid by substitution and unique by an easy argument like this one.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.