How would I integrate equations of the following form:

$$\frac{d a(t)}{dt}=ka(t)$$

where $k$ is constant.

I have the feeling that this is quite simple, but I seem to be stuck.

My initial thought was that I could do:

$$a(t) da(t) = k dt \iff \frac{a^2(t)}{2} = kt + c $$

But for some reason I don't think this is right. Should I instead do integration by parts as done in Integrating an unknown function?

  • 3
    $\begingroup$ $\frac{da(t)}{a(t)}=k\,dt$ could be better $\endgroup$ – Claude Leibovici Jun 25 '20 at 10:27

This is quite an infamous differential equation, one of the basic ones you learn in calculus courses. Use separation of variables. $$ \frac{a(t)}{dt} = ka(t) \implies \frac{da(t)}{a(t)}= kdt \implies \int \frac{da(t)}{a(t)}=\int k dt $$ and so on.

  • $\begingroup$ I multiplied instead of dividing out of pure distraction. Thank you so much for your help. $\endgroup$ – user7077252 Jun 25 '20 at 10:33
  • $\begingroup$ @user7077252 No problem! Accept the answer if it was what you were looking for :) $\endgroup$ – Omega Jun 25 '20 at 10:35
  • $\begingroup$ Only allowed to do so in a minute's time $\endgroup$ – user7077252 Jun 25 '20 at 10:37

The general solution to this ODE is $a(t) = Be^{kt}$, $B \in \Bbb{R}$. Note that you made an algebraic error. On the LHS, it should be $\dfrac{da}{a}$, so upon integration you should get $\ln|a| = kt +c$.


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