Setting up a differential equation with interest 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?
 A: Suppose the initial principal is $A$, and $12,000$ is withdrawn at the end of each year.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More generally, if an amount $K$ is withdrawn at the end of each of $N$ years, and the annual interest rate is $r$, the initial principal must be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.
A: 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
$$
            e^{-0.05(20)}A(20)-A(0) = -12,000\frac{1}{-0.05}(e^{-0.05(20)}-1),\\
                  -P = 240,000(e^{-1}-1),\\
                   P = 240,000(1-1/e).
$$
