I have the following ODE yielded as a steady state solution to a more complicated ODE. The important part is just to consider the following:

$$d/dt \langle N(t) \rangle=k-\Gamma \langle N(t) \rangle$$ where $\langle N(t) \rangle$ can be seen as $y$ or an average of $y$ (brackets indicating avg). I am not sure if that is the reason for the result.

The solution is given by $$\langle N(t) \rangle=k/\Gamma (1-e^{-\Gamma t})+\langle N(t) \rangle e^{-\Gamma t}$$

I do not see that however. So far, I've solved first for $$d/dt <N(t)>=\Gamma <N(t)>=0$$

Yielding $$\langle N(t) \rangle>=C\cdot e^{-\Gamma t}$$

and $C$ $$C=k/\Gamma (1-e^{-\Gamma t})+c_0$$

I do not see how it solves the equation by inserting c.

Any help is highly appreciated :)

edit: I had a thought it may have to do with the average being additive, and starting and $\langle N(0) \rangle$ being the same as $\langle N(t) \rangle$ since it is steady state. I am not completely sure however.

  • $\begingroup$ Use \langle and \rangle for angle brackets: $\langle$ $\rangle$ $\endgroup$
    – Dylan
    Commented Jul 20, 2018 at 18:43
  • $\begingroup$ What is the $\Gamma$? $\endgroup$
    – Botond
    Commented Jul 20, 2018 at 18:52
  • $\begingroup$ $\Gamma$ is just a constant :) $\endgroup$
    – user469216
    Commented Jul 20, 2018 at 18:53

1 Answer 1


Let $y = \langle N(t)\rangle$, so that we may write $$\frac{dy}{dt} +\Gamma y = k$$ This equation can be solved by method of integrating factors. An integrating factor here is $e^{\int_0^t \Gamma\, ds} = e^{\Gamma t}$. Multiply the equation by $e^{\Gamma t}$ and note that by the product rule, we obtain $$\frac{d}{dt}(e^{\Gamma t} y) = ke^{\Gamma t}$$ Integrate both sides to get $$e^{\Gamma t}y = \frac{k}{\Gamma}e^{\Gamma t} + C$$where $C$ is a constant. Setting $t = 0$, $y(0) = k/\Gamma + C$, so then $C = y(0) - k/\Gamma$. Therefore $$y = \frac{k}{\Gamma} + \left(y(0) - \frac{k}{\Gamma}\right)e^{-\Gamma t}= \frac{k}{\Gamma}\left(1 - e^{-\Gamma t}\right) + y(0)e^{-\Gamma t}$$ as desired.


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