Question

Consider a loan repayment plan described by the initial value problem $$B^{\prime}(t)=0.03 B-600, \quad B(0)=40,000$$ where the amount borrowed is \(B(0)=\$ 40,000,\) the monthly payments are \(\$ 600,\) and \(B(t)\) is the unpaid balance in the loan. a. Find the solution of the initial value problem and explain why \(B\) is an increasing function. b. What is the most that you can borrow under the terms of this loan without going further into debt each month? c. Now consider the more general loan repayment plan described by the initial value problem $$B^{\prime}(t)=r B-m, \quad B(0)=B_{0}$$ where \(r>0\) reflects the interest rate, \(m>0\) is the monthly payment, and \(B_{0}>0\) is the amount borrowed. In terms of \(m\) and \(r,\) what is the maximum amount \(B_{0}\) that can be borrowed without going further into debt each month?

Short answer

Answer: The maximum initial amount B_0 that can be borrowed without going further into debt each month is given by B_0 = m/r, where m is the monthly payment amount, and r is the monthly interest rate.

Step-by-step solution

Separate variables in the differential equation

First, we need to rewrite the differential equation as a product of a function of \(t\) and a function of \(B\). This can be done as: $$\frac{dB}{dt} = 0.03B - 600 \Rightarrow \frac{dB}{0.03B - 600} = dt$$

Integrate both sides

Now, we integrate both sides with respect to their variables: $$\int \frac{dB}{0.03B - 600} = \int dt$$ This results in: $$\frac{1}{0.03}\ln|0.03B - 600| = t + C$$

Solve for B(t)

Exponentiate both sides to remove the natural logarithm, and then solve for \(B(t)\): $$0.03B - 600 = Ce^{0.03t}$$ $$B(t) = \frac{Ce^{0.03t}+600}{0.03}$$

Apply the initial condition

To find the particular solution, apply the initial condition \(B(0) = 40000\): $$B(0) = \frac{Ce^{0.03\cdot 0}+600}{0.03} = \frac{C+600}{0.03} = 40000$$ $$C = 0.03\cdot 40000 - 600 = 600$$ So, the particular solution is: $$B(t) = \frac{600(e^{0.03t}+1)}{0.03}$$ #a. Explaining why B(t) is an increasing function#

Examine the rate of change

Now we need to determine if \(B(t)\) is an increasing function. This will be true if its derivative, \(B'(t)\), is positive for all values of \(t\). From the original differential equation: $$B'(t)=0.03 B(t)-600$$ We substitute the expression for \(B(t)\) we found earlier: $$B'(t)=0.03 (\frac{600(e^{0.03t}+1)}{0.03})-600$$ $$B'(t)=600(e^{0.03t}+1)-600$$ Since \(e^{0.03t}\) is always positive and increases with \(t\), the function \(B'(t)\) is also positive and increases with \(t\). Therefore, \(B(t)\) is increasing. #b. Determining the maximum loan amount#

Find when B'(t) is non-negative

To find when \(B'(t)\) is non-negative, we set it to zero and solve for \(B(t)\): $$600(e^{0.03t}+1)-600 = 0$$ $$e^{0.03t}+1=1$$ $$e^{0.03t}=0$$ However, the exponential function \(e^{0.03t}\) can never be equal to zero. Therefore, there is no maximum loan amount that can be borrowed under the terms of this loan without going further into debt each month. #c. Analyzing the general loan repayment plan#

Examine B'(t) in general case

The general differential equation is given by: $$B'(t) = rB - m$$ To find the maximum amount that can be borrowed without going further into debt each month, we need to find when \(B'(t)\) is non-negative. We set \(B'(t)\) to zero and solve for \(B\): $$rB - m = 0$$ $$B = \frac{m}{r}$$ So, the maximum initial amount \(B_0\) that can be borrowed without going further into debt each month is given by: $$B_0 = \frac{m}{r}$$

Key concepts

Loan Repayment

A loan repayment plan describes how a borrower will pay back a loan over time. This includes both the total amount borrowed, known as the principal, and the interest that accrues over time.
In our original exercise, the loan repayment plan is modeled using a differential equation. The equation captures the dynamics of the unpaid balance on a loan through time, incorporating factors like fixed monthly payments and interest rates.
For example, in the exercise, the monthly payment is $600 while the borrowed amount, or the initial loan balance, is $40000. The differential equation describes how this balance changes every month as payments are made and interest accumulates.

Initial Value Problem

An initial value problem in mathematics involves a differential equation along with a specified initial condition. This condition helps to find a particular solution to the differential equation.
In the context of loan repayment, the initial value problem outlines how the loan balance will change from its initial state over time. The given conditions specify not just the differential equation itself, but also the initial loan amount. When the exercise specifies that \( B(0) = 40000 \), it means that the initial amount borrowed is $40000. The initial value acts as a starting point that determines the unique path of the function that solves the differential equation, giving us more accurate information about the loan balance at any given time.

Interest Rate

Interest rate is a crucial concept in any loan repayment scenario. It represents the cost of borrowing money from a lender, expressed as a percentage of the loan balance that the borrower must pay at regular intervals.
In the differential equation from the exercise, the interest rate is symbolized by the coefficient 0.03. This means that the loan accrues interest at a rate of 3% per month on the remaining unpaid balance.

• This rate influences how quickly the overall balance can grow or shrink, based on whether the interest charge exceeds the monthly payment.
• If interest charged is greater than what is repaid, the total unpaid loan balance grows.
• If the monthly payment exceeds the interest charged, the outstanding balance decreases over time.

Exponential Function

The exponential function plays a vital role in solving differential equations, especially those modeling real-world financial scenarios like loans. It captures the way quantities grow or decrease at a consistent percentage rate.
In solving the original problem, we arrived at an expression involving the exponential function \( e^{0.03t} \). This function reflects how the balance of the loan evolves as interest continually compounds over time.

• The presence of \( e^{0.03t} \) indicates continuous growth due to the compounding nature of interest.
• The exponential function inherently displays growth because the exponential of any real number is always positive.
• This property helps us conclude whether the loan balance is increasing or decreasing over time.

Understanding exponential growth is key in appreciating why loan balances could become larger if not carefully managed.