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

In the given initial value problem, the solution is found to be \(B(t) = -20000e^{0.03t} + 60000e^{-0.03t}\). It is shown that B(t) is an increasing function as its derivative B'(t) is positive. The maximum borrowed amount without going further into debt under the terms of the loan is $20,000. In terms of the general loan repayment plan, the maximum amount B₀ that can be borrowed without going further into debt each month is given by \(B_{0} = \frac{m}{r}\), where m represents the monthly payment and r represents the interest rate.

Step-by-step solution

Find the solution of the initial value problem

For the given initial value problem, we have the differential equation and the initial condition: $$B^{\prime}(t)=0.03 B-600, \ B(0)=40,000$$ This is a first order linear differential equation. The standard form is: $$B^{\prime}(t) - P(t)B = Q(t)$$ In our case, we have P(t)=0.03 and Q(t)=-600. To solve this differential equation, we first need to find the integrating factor e^(∫P(t)dt) which is: $$e^{\int0.03dt}=e^{0.03t}$$ Now, we multiply both sides of the equation by the integrating factor: $$e^{0.03t}B^{\prime}(t) - 0.03 e^{0.03t}B = -600e^{0.03t}$$ The left side of the equation is now the derivative of the product: $$\frac{d}{dt}[Be^{0.03t}] = -600e^{0.03t}$$ Integrating both sides with respect to t: $$\int d[Be^{0.03t}] = -600\int e^{0.03t}dt$$ Integrating and solving for B(t): $$Be^{0.03t} = -600 \frac{1}{0.03} e^{0.03t} + C$$ $$B(t) = -20000e^{0.03t} + Ce^{-0.03t}$$ Now, we use the initial condition to find C: $$B(0) = -20000e^{0} + Ce^{0} = 40000$$ $$C = 60000$$ Therefore, the solution of the initial value problem is: $$B(t) = -20000e^{0.03t} + 60000e^{-0.03t}$$

Explain why B(t) is an increasing function

To show that B(t) is an increasing function, we need to show that its derivative B'(t) is positive: $$B^{\prime}(t)=0.03[-20000e^{0.03t} + 60000e^{-0.03t}] -600$$ Dividing by 60000 (which doesn't affect the sign): $$\frac{B^{\prime}(t)}{60000}=-\frac{1}{3}e^{0.03t} + e^{-0.03t} - \frac{1}{100}$$ Now, we need to show that this expression is positive. Observe that e^(0.03t) ≥ 1 and e^(-0.03t) ≥ 1. In that case: $$-\frac{1}{3}e^{0.03t} + e^{-0.03t} - \frac{1}{100} \geq - \frac{1}{3}(1) + 1 - \frac{1}{100}$$ $$ \geq - \frac{1}{3} + 1 - \frac{1}{100} = \frac{94}{300} > 0$$ Thus, B'(t) is positive, and B(t) is an increasing function.

Find the most you can borrow without going further into debt

To identify the maximum borrowed amount without going further into debt, we need to find when B'(t) will be equal to 0: $$0.03 B - 600 = 0$$ Solving for B: $$B = \frac{600}{0.03} = 20,000$$ Thus, the most we can borrow under the terms of this loan without going further into debt each month is $20,000.

Find the maximum B₀ that can be borrowed in terms of m and r

For the general loan repayment plan, we have: $$B^{\prime}(t)=r B - m, \ B(0)=B_{0}$$ To find the maximum borrowed amount without going further into debt, we need to find when the derivative B'(t) will be equal to 0: $$rB - m = 0$$ Solving for 𝐵₀: $$B_{0} = \frac{m}{r}$$ Therefore, the maximum amount 𝐵₀ that can be borrowed without going further into debt each month in terms of m and r is given by \(B_{0} = \frac{m}{r}\).

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation problem in which the solution is needed for a specific intial condition. This means you're seeking a function that not only satisfies the differential equation but also matches a set starting point or initial value.
In our example, the initial value problem is described by the equation:

• \(B^{\prime}(t)=0.03 B-600\)
• The initial condition is \(B(0)=40,000\).

To solve it, identify the necessary conditions and ensure that the function discovered meets them. The initial value provides a starting point, allowing you to determine constants during the integration process.

Loan Repayment

When dealing with loans, repayment is a critical aspect. Loan repayment typically involves paying small sums over time to clear a large borrowed amount. The goal is to pay back the loan amount before its deadline.
In our exercise, the loan repayment involves an initial borrowed amount of \(\\(40,000\) with monthly payments of \(\\)600\). This repayment plan will affect the balance owed over time. It is crucial to manage monthly payments effectively so that the borrower does not fall into more debt as time goes on.

Interest Rate

The interest rate is an added percentage that lenders charge borrowers for using their money. It impacts how quickly or slowly the loan balance changes over time.
In the formula \(B^{\prime}(t)=r B - m\), \(r\) represents the interest rate.

• A positive \(r\) implies an increase in the debt over time if no payments are made.
• It's important for the interest rate to be balanced with the repayment amount to prevent the loan from becoming overwhelming.

In the initial problem scenario, a 3% interest rate influences the unpaid balance by increasing the amount owed, requiring effective monthly payments to offset this increase.

Increasing Function

Understanding why a function is increasing is important for predicting how values will change over time. An increasing function is one where, as \(t\) (time) increases, so does \(B(t)\) (the balance in our case).
In this case, if \(B^{\prime}(t) > 0\), that indicates \(B(t)\) is increasing. This can be verified by checking the behavior of the derivative function.

• Given: \(B^{\prime}(t)=0.03 B-600\)
• For \(B(t)\) to be increasing, ensure \(-\frac{1}{3}e^{0.03t} + e^{-0.03t} - \frac{1}{100} > 0\).

These checks confirm the overall direction that the function \(B(t)\) takes over time, which, in this exercise, shows that the balance could increase over time under certain conditions.