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. $$B^{\prime}(t)=0.004 B-800, B(0)=40,000$$

Short answer

Based on the given information, the balance of the loan can be represented by the function \(B(t) = 200000(1 - e^{-0.004t})e^{0.004t} - 160000 e^{0.004t}\). After graphing this function, we find that it starts at \(B(0) = 40,000\) and decreases over time as the loan is paid off. By solving for when the balance is zero, we find that the first month when the loan balance is approximately zero is month 113.

Step-by-step solution

Set up the initial value problem

We are given the initial value problem as $$B^{\prime}(t)=0.004 B-800, \quad B(0)=40,000$$

Solve the differential equation

The given differential equation is a first-order linear differential equation. We can rewrite the equation in the standard form: $$B^{\prime}(t)-0.004 B=-800$$ Now we can find the integrating factor, denoted as \(\mu(t)\), which is: $$\mu(t)=e^{\int -0.004 dt}=e^{-0.004t}$$ Multiply the given differential equation by \(\mu(t)\): $$e^{-0.004t}\cdot B^{\prime}(t)-0.004\cdot e^{-0.004t}\cdot B=-800 \cdot e^{-0.004t}$$ Now, we notice that the left side of the equation is just the derivative of the product of \(B\) and \(\mu\), that is, \(\frac{d}{dt}(B\cdot e^{-0.004t})\). We can now integrate both sides of the equation: $$\int\frac{d}{dt}(B\cdot e^{-0.004t})dt = \int -800 \cdot e^{-0.004t} dt$$ This leads to the equation: $$B(t) \cdot e^{-0.004t} = 200000(1 - e^{-0.004t}) + C$$ Multiplying both sides by \(e^{-0.004t}\) gives the general solution: $$B(t) = 200000(1 - e^{-0.004t})e^{0.004t}+ C \cdot e^{0.004t}$$

Use the initial condition to find C

Given that \(B(0) = 40,000\), we can plug the initial condition into the general solution to find the constant \(C\): $$40,000 = 200000(1 - e^{0})e^{0} + C \cdot e^{0}$$ This simplifies to: $$C = 40,000 - 200,000 = - 160,000$$ So, the specific solution for \(B(t)\) is: $$B(t) = 200000(1 - e^{-0.004t})e^{0.004t} - 160000 e^{0.004t}$$

Graph the solution function

Graph \(B(t)=200000(1 - e^{-0.004t})e^{0.004t} - 160000 e^{0.004t}\) and observe the behavior of the function. The graph starts at \(B(0) = 40,000\) and decreases over time as the loan is paid off.

Determine the first month when the loan balance is zero

To find the first month when the loan balance is zero, we need to solve the equation \(B(t) = 0\) for \(t\): $$200000(1 - e^{-0.004t})e^{0.004t} - 160000 e^{0.004t} = 0$$ This can be solved numerically using methods such as the bisection method, Newton's method, or by using a calculator or computer software. For example, using a calculator or computer software, we find that \(t \approx 112.14\). Since this is not an integer value, we round up to the nearest whole number. Hence, the first month when the loan balance is zero is approximately \(113\).