I need help with the following ODE: $$ \frac{\partial S}{\partial t} - (a+b) S = -A(t)$$

The solution is supposed to be:

$$S(t) = \int_{t}^{\infty}A(u) \> e^{-(a+b)(u-t)}du.$$

The integrating factor I calculated is:

$M(u) = e^{-(a+b)u}$, since it solves the equation: $-(a+b) M(u) = M'(u)$.

I struggle when I use the reverse product rule: $$\frac{d}{dt} \bigl(S(t)M(u)\bigr) = -A(t)M(u)$$ and integrate afterwards: $$S(t)M(u) = \int -A(t) M(u) du.$$ Then divide by $M(u)$. $$S(t) = \int -A(t)M(u) du \> \cdot e^{(a+b)u} $$

Where will the sign of A change? Further, what's the right auxiliary variable to use here, since there are function dependent on t and u will in the end be the auxiliary variable, but I don't know what's the correct variable to use for differentiation and integration, I guess $u$ and $t$ could be used interchangeably. Additionally, when do I need to insert the limits for the integral?

PS. Could I eventually simply replace $u$ by $t$ in the second factor on the right-hand side, such that it can be multiplied with the function under the integral sign?


1 Answer 1


Note that the integral in the solution is from $t$ to $+\infty$, so $$[S(t)e^{-(a+b)u}]_t^{+\infty} = \int_t^{+\infty} -A(u) e^{-(a+b)u} du$$ implies $$0-S(t)e^{-(a+b)t} = \int_t^{+\infty} -A(u) e^{-(a+b)u} du$$ and $$S(t) = \int_{t}^{+\infty}A(u) e^{-(a+b)(u-t)}du.$$

  • $\begingroup$ Since, I didn't know when to implement the limits to the integral function that is very helpful. So what I missed was to evaluate the left-hand side at the limits, right? $\endgroup$
    – Sam
    Commented Dec 18, 2018 at 12:50
  • $\begingroup$ Yes, here they are interested in the solution which goes to zero as $t$ goes to $+\infty$ $\endgroup$
    – Robert Z
    Commented Dec 18, 2018 at 12:58

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