# Help to solve this ODE by an integrating factor

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?

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 $$S(t) = \int_{t}^{+\infty}A(u) e^{-(a+b)(u-t)}du.$$
• Yes, here they are interested in the solution which goes to zero as $t$ goes to $+\infty$ Commented Dec 18, 2018 at 12:58