Example 4 from an old set of MIT notes asks:
Bank interest. On a savings account, a bank pays the continuous interest rate $r$, meaning that a sum $A_0$ deposited at time $u = 0$ will by time $u = t$ grow to the amount $A_0e^{rt}$.
Suppose that starting at day $t = 0$ a Harvard square juggler deposits every day his take, with deposit rate $d(t)$ — i.e., over a relatively small time interval $[u_0, u_0 +∆u]$, he deposits approximately $d(u_0)∆u$ dollars in his account. Assuming that he makes no withdrawals and the interest rate doesn’t change, give with reasoning an approximate expression (involving a convolution) for the amount of money in his account at time $u = t$.
I'm confused. I know that the convolution formula is $f(t) * g(t) = \int_0^t f(t)\ g(t - u)\ du$, but I'm not sure if these are the same $t$ and $u$ as in the problem (in my notes, $u$ is a dummy variable, while in the problem, it seems to be a second time variable). I also know that the full solution to an IVP is the sum of a state-free solution and an input-free solution, the former given by $y_s = y_\delta * d(t)$. For the latter, I only have the second-order formula $y_i = y_\delta'y(0) + d_\delta y'(0) + ay_\delta y(0)$.
Using the Laplace Transform, $A_\delta ' = rA_\delta + \delta, A_\delta(0) = 0 \implies sℒ(A_\delta) = rℒ(A_\delta) + 1 \implies ℒ(A_\delta) = \frac{1}{s - r} \implies A_\delta = e^{rt}$. Therefore, the state-free solution should be $A_s = e^{rt} * d(t)$. The second order formula for the input-free solution won't work here (no $A'(0)$ is given) and I can't find a first-order equivalent. Even so, I'm not sure if I'm on the right track as there seems to be a lot of information given in the problem that my solution doesn't utilize. How is this problem to be solved?