Adam opens a superannuation account with a deposit of $1000$ and intends to deposit money at a rate of $1000$ per year for the next 25 years. Interest is paid at the rate of $0.1$% pa, compounded continuously. If A is the amount after $t$ years, prove that $$A = 11000e^{0.1t}-10000$$
I simply have $A = 1000e^{0.1t}$. Obviously this is wrong but I cant see where or why anything would be subtracted in this case. To try and account for the deposits, I set up a recursive balance which I found the sum of the geometric progression to be $A = 1000\left[\frac{(e^{0.1})^{t+1} -1}{e^{0.1} -1}\right]$
It cam from $A_0=1000$
$A_1 = 1000e^{0.1}+1000$...
. . . $A_t =1000\left[\frac{(e^{0.1})^{t+1} -1}{e^{0.1} -1}\right]$
Please help
Let $A(\tau)$ denote the balance at time $\tau$.
Over the interval $[\tau, \tau+ \Delta \tau]$ we have
$$A(\tau + \Delta \tau ) \approx A(\tau) (1 + i \Delta \tau) + \alpha \Delta \tau,$$
where $i$ is the interest rate and $\alpha$ is the deposit rate (per annum).
Rearranging we get
$$\frac{A(\tau + \Delta \tau ) - A(\tau)}{\Delta \tau} = i A(\tau) + \alpha$$
Taking the limit as $\Delta \tau \to 0$, we get the differential equation for the account balance
$$\frac{dA}{d \tau} = i A + \alpha$$
With the initial balance $A(0) = A_0$ we solve by separation of variables to obtain
$$\int_{A_0}^{A(t)}\frac{d A}{\alpha + iA} = \int_0^t \,d \tau, \\\implies \frac{1}{i}\log \frac{\alpha +i A(t)}{\alpha + i A_0} = t, \\ \implies A(t) = (\alpha /i + A_0)e^{it} - \alpha/i$$
Substituting parameters, we obtain
$$A(t) = 11000e^{0.1t}- 10000$$