Setting up a differential equation with interest

Question

The following is the problem that I am working on.

A bank offers an annual interest of 5% per year with continuous interest for 20 yrs. Every year, a person withdraws 12,000 from the account. If the balance equals exactly 0 at the end of the 20th year, what is the principal of the account?

To set up the differential equation is the part that I am not confident.

Intuitively speaking, I know the continuous interest formula would be

$$A=Pe^{.05t}$$

and every year the person uses $12,000 from the account so the total usage is

$$C=-12,000t$$

after t years.

So I am thinking that the differential equation should have a solution that looks something like

$$B=Pe^{.05t}-12,000t$$

and to set this equation as a differential equation, I let

$$\frac{dB}{dt}=\frac{dA}{dt}+\frac{dC}{dt}$$

$$B'=A'-C'$$

$$B'=(1.05)A-12,000$$

But I am not confident if my setup is correct. Can someone lend a hand?

Answer 1

Suppose the initial principal is $A$, and $12,000$ is withdrawn at the end of each year.

At the end of the first year (immediately after the withdrawal), $A(1.05)^1-12,000$ remains.

This grows for another year, and then the second withdrawal is made. At the end of the second year, $A(1.05)^2-12,000(1.05) - 12,000$ remains.

Continuing, one sees that at the end of twenty years, since the balance is zero, we have $$0 = A(1.05)^{20} - 12,000\left(1 + (1.05)^1+(1.05)^2+\cdots + (1.05)^{19}\right)$$

The geometric sum can be simplified as $\frac{1-1.05^{20}}{1-1.05} = \frac{1.05^{20} - 1}{0.05}$, so that $$A = 12,000\left(\dfrac{1.05^{20}-1}{0.05}\right)(1.05^{20}) = \boxed{12,000\left(\dfrac{1 - 1.05^{-20}}{0.05}\right) \approx 149,546.52}$$

More generally, if an amount $K$ is withdrawn at the end of each of $N$ years, and the annual interest rate is $r$, the initial principal must be

$$K\left(\dfrac{1 - (1+r)^{-N}}{r}\right)$$

Note: I'm assuming you mean for the effective annual rate to be $5$%? If not, you can convert the continuous compounding rate to an effective annual rate using $1+r = e^{0.05}$ and correct the answer accordingly. I am using effective rate in the general formula.

Answer 2

Let $A(t)$ be the amount in the bank as a function of $t$ in years. Then $$ \frac{dA}{dt} = 0.05A -12,000,\\ A(0)=P \mbox{ is the principal},\\ A(20)=0. $$ The integrating factor is $e^{-0.05t}$, which gives $$ \frac{d}{dt}(e^{-0.05t}A) = -12,000e^{-0.05t}. $$ Integrating from $t=0$ to $t=20$ gives $$ e^{-0.05(20)}A(20)-A(0) = -12,000\frac{1}{-0.05}(e^{-0.05(20)}-1),\\ -P = 240,000(e^{-1}-1),\\ P = 240,000(1-1/e). $$