Science:Math Exam Resources/Courses/MATH105/April 2014/Question 05 (a)

MATH105 April 2014

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Question 05 (a)

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem $B'(t)=aB-m$ for $t\geq 0$, with $B(0)=B_{0}$. The constant $a$ reflects the annual interest rate, $m$ is the annual rate of withdrawal, and $B_{0}$ is the initial balance in the account. Solve the initial value problem with $a=0.02$ and $B(0)=B_{0}=\$30,000$. Note that your answer depends on the constant $m$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Separation of variables.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We first solve for the differential equation: ${\begin{aligned}{\frac {dB}{dt}}&=aB-m\\dB&=(aB-m)dt\\{\frac {dB}{aB-m}}&=dt\quad ({\text{separating the variables}})\\\int {\frac {dB}{aB-m}}&=\int dt\quad \\{\frac {1}{a}}\ln |aB-m|&=t+C\\\ln |aB-m|&=at+C\quad ({\text{or aC, but C is arbitrary so we still call it C}})\\|aB-m|&=e^{at+C}=Ae^{at}\quad ({\text{where }}A=e^{C})\\aB-m&=\pm Ae^{at}\quad ({\text{removing the absolute values gives a }}\pm )\\B&={\frac {1}{a}}(m-Ae^{at})\quad ({\text{A here is }}\pm A{\text{ is arbitrary and still unknown}})\end{aligned}}$ Now we can use the initial conditions with $B(0)=30000$ and $a=0.02=1/50$ to find: $B(0)=30000=50(m-A)\implies A=m-600$ and thus $B(t)=50(m-(m-600)e^{0.02t})=50((600-m)e^{0.02t}+m)$.

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