Science:Math Exam Resources/Courses/MATH105/April 2010/Question 01 (k)/Solution 1

The instantaneous rate of change of capital has two components:

• We continuously add money at a rate of $3,000 per year;
• We earn 6% interest compounded continuously.

If we denote by A(t) the balance of the account at time t, then we can write the differential equation

${\begin{aligned}{\frac {dA}{dt}}&=3000+6\%\,A(t)\\&=3000+0.06\,A(t)\end{aligned}}$

We solve this equation using separation of variables:

$\int {\frac {dA}{3000+0.06A}}=\int \,dt$

Solving both sides separately yields

${\frac {1}{0.06}}\ln |3000+0.06A|+C_{1}=t+C_{2}$

Since A is positive, as the amount of money in the account, we can drop the absolute value sign. This yields

$\displaystyle \ln(3000+0.06A)=0.06t+C_{3}$

or, equivalently

$3000+0.06A=e^{0.06t+C_{3}}=e^{0.06t}e^{C_{3}}=C_{4}e^{0.06t}.$

C₄ is another arbitrary, but positive constant. Hence we obtain for A(t):

$A(t)={\frac {1}{0.06}}\left(C_{4}e^{0.06t}-3000\right).$

Now it's time to think about the initial condition. Since the account does not start with any balance initially, we know that A(0) = 0. This yields for C₄ that

$0={\frac {1}{0.06}}\left(C_{4}e^{0}-3000\right),\quad {\text{which implies}}\quad C_{4}=3000.$

With this we finally find the expression for the balance in dollars at t years:

$A(t)={\frac {3000}{0.06}}\left(e^{0.06t}-1\right).$

This implies that at t = 5 years the balance is

$A(5)={\frac {3000}{0.06}}\left(e^{0.06\cdot 5}-1\right)=17492.94$

Reality check: Without interest we would expect to see 5 x 3000 = 15 000 dollars in the account. The amount with interest should be in the same order of magnitude and a little larger.