Question

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$B^{\prime}(t)=0.0075 B-1500, B(0)=100,000$$

Short answer

Answer: The loan balance will be zero in the 4th month.

Step-by-step solution

Write the initial value problem

We have the initial value problem given by the first-order linear differential equation: $$B^{\prime}(t)=0.0075 B-1500$$ with the initial condition: $$B(0)=100,000$$

Solve the first-order linear differential equation

To solve the given first-order linear differential equation, we can use an integrating factor. The integrating factor is given by: $$e^{\int 0.0075 dt} = e^{0.0075t}$$ We multiply both sides of the differential equation by the integrating factor to obtain: $$e^{0.0075t}B^{\prime}(t) + 0.0075e^{0.0075t}B = 1500e^{0.0075t}$$ Now, the left side of this equation is the derivative of the product \(e^{0.0075t}B(t)\). So, we integrate both sides with respect to \(t\): $$\int (e^{0.0075t}B^{\prime}(t) + 0.0075e^{0.0075t}B)dt = \int 1500e^{0.0075t} dt$$ This yields: $$e^{0.0075t}B(t) = 200,000e^{0.0075t} - C$$ Now, we solve for \(B(t)\): $$B(t) = 200,000 - Ce^{-0.0075t}$$

Use initial condition to find the constant C

Now we will use the given initial condition, \(B(0)=100,000\), to find the constant \(C\): $$100,000 = 200,000 - Ce^{0}$$ Solving for \(C\), we get: $$C = 100,000$$ Our final solution for the balance \(B(t)\) is: $$B(t) = 200,000 - 100,000e^{-0.0075t}$$

Find the first month when the loan balance is zero

To find the first month when the loan balance is zero, we set \(B(t) = 0\) and solve for \(t\): $$0 = 200,000 - 100,000e^{-0.0075t}$$ Solving the equation for \(t\), we get: $$e^{-0.0075t} = 2$$ Taking the natural logarithm of both sides, we obtain: $$-0.0075t=\ln(2)$$ Dividing by \(-0.0075\), we get: $$t=\frac{\ln(2)}{-0.0075}$$ Don't forget, since the question asks for months, that's what we need to find. First, calculate t: $$t \approx 92.3$$ Now, divide by 30 to convert days into months: $$months \approx \frac{92.3}{30} \approx 3.08$$ Thus, the first month when the loan balance is zero is approximately in the 4th month (since it cannot be the 3rd month).

Key concepts

Differential Equations

Differential equations are fundamental to modeling various real-world processes, and they describe how things change. In our loan payoff problem, we're dealing with a first-order linear differential equation, which has the form \( B^{\prime}(t) = kB + c \). Here, \( B^{\prime}(t) \) is the rate of change of the loan balance over time, \( k \) is the rate at which it grows or decays, and \( c \) represents constant payments applied to reduce the balance.
The equation's solution gives us a function that describes how the balance of the loan changes over time. Solving it involves understanding the components that drive the rate of change, such as the interest (in this case represented by the constant \( 0.0075 \)) and the regular payment made to decrease the loan balance (\( -1500 \)).

Initial Value Problem

An initial value problem is a scenario where the initial state of the system is known and one seeks to determine how this state evolves over time. In our exercise, the initial condition is given by \( B(0) = 100,000 \). This tells us the loan balance immediately at the start — or at time \( t = 0 \).
With this initial condition, the specific solution to the differential equation can be found. It defines a unique curve within a family of solutions because any given differential equation with different initial conditions could have different solutions.

• This ensures that we can reference a specific timeline or balance change based on the initial loan balance.
• It gives context to the solution, helping us tailor predictions or analyses specific to our scenario.

Integrating Factor

The integrating factor method is a useful technique for solving linear differential equations. By multiplying the entire differential equation by an integrating factor, we transform it into a form that can be easily integrated. The integrating factor is generally of the form \( e^{\int P(t) \, dt} \).
In our loan modeling exercise, we used \( e^{0.0075t} \) as the integrating factor, which helped us to arrange the equation conveniently to find a solution.

• This allows us to take complex equations and consolidate them into integrated forms.
• The left side of the equation becomes the derivative of a product, simplifying the integration process.

Once the equation is transformed, integrating both sides unlocks the solution we need for the loan balance function, \( B(t) \).

Exponential Growth and Decay

Exponential growth and decay refer to the continuous change of a quantity according to a fixed percentage growth or shrink per unit time. This principle is directly applied to finance, especially in scenarios involving interest rates.
In our loan payoff modeling, the loan balance — affected by constant payments — decays over time. The equation \( B(t) = 200,000 - 100,000e^{-0.0075t} \) beautifully illustrates decay, where the exponential function \( e^{-0.0075t} \) models the decrease in balance as time progresses.

• Every month, the balance shrinks, moving closer to zero, thanks to payments exceeding the compound interest.
• The effectiveness of the payment against interest is clear and can be plotted over time to visualize the "decay".

This modeling helps borrowers understand how long it takes to completely pay off a loan, offering insights into strategic financial planning.