An investor deposits USD 300 in a bank account at time 0, reinvests all interest payments and also additionally continuously invests USD 300 per annum, until the total value of the deposits reaches USD 3312. At the point the investor stops making additionally deposits, but still lets the interest payments accumulate in the account.
The ODE for the value of the deposits, $V$, over time is then, $$\frac{dV}{dt}=r(t)V(t)+I(t),$$ where $I(t)=300$ until $V(t)$ reaches $V$=3312, at which point $I(t)$ instantaneously switched to $I(t)=0.$
Derive an expression for the value of the asset as a function of time, $V(t)$, $t\geq 0.$
I tried to work through this problem, but I am not quite sure how can I formulate it correctly. I tired several approaches, but can't seem to get my head around this problem. Any help will be greatly appreciated. Thanks in advance!
Let be $V_0=300$ the deposit at $t=0$, $V(T)=3312$ the value at $t=T$, and lest's assume the interest rate $r(t)=r$ to be constant. Denote $$I(t)=\rho\left(1-H(t-T)\right)$$ the rate of payment at any time in the range $[0,T]$ where $\rho=300$ and $H(t)$ is the Heaviside function or step function defined as $$H(t)=\begin{cases} 1 & t> 0\\ 0 & t\le 0 \end{cases}$$ so that $$I(t)=\begin{cases} 0 & t> T\\ \rho & 0\le t\le T \end{cases}$$ So you have to solve $$ \frac{\mathrm d V}{\mathrm d t}=r V(t)+I(t)\tag 1 $$ that is a first order non-linear ODE. It's better to solve the equation (1) separately in the time intervals $t\le T$ and $t>T$ $$ \begin{align} \frac{\mathrm d V}{\mathrm d t}&=r V(t)+\rho & 0\le t\le T\tag 2\\ \frac{\mathrm d V}{\mathrm d t}&=r V(t) & t> T\tag 3 \end{align} $$ The solution is easy to find and it is
You can check the result observing that the initial amount $V_0$ will produce $V_0\mathrm{e}^{rt}$ with a continuous compounding. The continuous payments with rate $\rho$ the value will produce for $0\le t \le T$ the value $$\int_0^t\rho\,\mathrm{e}^{rs}\mathrm{d}s=\rho\left[\frac{\mathrm{e}^{rs}}{r}\right]_0^t=\frac{\rho}{r}\left(\mathrm{e}^{rt}-1\right)$$ and then for $0\le t \le T$ we have $$ V(t)=V_0\mathrm{e}^{rt}+\frac{\rho}{r}\left(\mathrm{e}^{rt}-1\right)=\left(V_0+\frac{\rho}{r}\right)\mathrm{e}^{rt}-\frac{\rho}{r} $$ For $t=T$, the amount in the bank account will be $$V(T)=\left(V_0+\frac{\rho}{r}\right)\mathrm{e}^{rT}-\frac{\rho}{r}$$ and no payments will be mabe for $t>T$. So the amount $V(T)$ will produce $V(T)\mathrm{e}^{r(t-T)}$ with a continuous compounding for $t>T$. So we find $$ V(t)= \begin{cases} V_0\mathrm{e}^{r t}+\frac{\rho}{r} \left(\mathrm{e}^{r t}-1\right)& 0\le t\le T\\ V(T)\mathrm{e}^{r (t-T)} & t> T \end{cases} $$ that is exactly the $(4)$.