Integral equation $u(t)=f(t)+a\int_0^t u(s)ds\quad t\geq 0$

Question

Let $a\in\mathbb R$ and $f\colon [0,1]\to\mathbb R$ a continuous function. Solve the integral equation $$u(t)=f(t)+a\int_0^t u(s)\mathrm ds,\quad t\geq 0$$ and find an explicit formula for the solution. Thank you

Answer 1

One approach is to solve this when $f$ is differentiable, to find a formula for $u$ involving only $f$ (this can be done using integration by parts to get rid of $f'$), and to show that this formula also yields the unique solution when $f'$ does not exist.

Another approach is to consider the function $v=u-f$ and to note that $$ v(t)=g(t)+a\int_0^tv(s)\mathrm ds,\qquad g(t)=a\int_0^tf. $$ The function $g$ is $C^1$ hence one can solve this new integral equation the usual way, by differentiatig it. One gets $v'=g'+av$, hence $(\mathrm e^{-at}v(t))'=\mathrm e^{-at}g'(t)$, that is, $$ v(t)=\mathrm e^{at}\left(v(0)+\int_0^t\mathrm e^{-as}g'(s)\mathrm ds\right). $$ Since $v(0)=0$ and $g'=af$, this shows that $$ u(t)=f(t)+v(t)=f(t)+a\mathrm e^{at}\int_0^t\mathrm e^{-as}f(s)\mathrm ds. $$ Sanity check: The first approach described in this post yields the same formula.