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. $$B^{\prime}(t)=0.01 B-750, B(0)=45,000$$

Short answer

Short Answer: The loan balance will be zero after approximately 78 months. To find this answer, we solved the given first-order linear ODE using the integrating factor method and applied the initial condition. The specific solution is \(B(t) = e^{0.01t} \left(\int -750e^{-0.01t} dt + 45,000\right)\). By setting the balance to zero and solving for \(t\), we found the first month when the balance reaches zero. The graph of this function would show a decreasing balance over time until it reaches zero at \(t = 78\) months.

Step-by-step solution

Identify the type of ODE

The given ODE is a first-order linear ODE since it can be written as: \(B^{\prime}(t) + (-0.01)B = -750\).

Solve the ODE using integrating factor

To solve a first-order linear ODE in the form \(B^{\prime}(t) + p(t)B = q(t)\), we can use the integrating factor method. The integrating factor (IF) is given by \(IF=e^{\int p(t) dt}\). For our problem, \(p(t) = -0.01\). Therefore, the integrating factor can be calculated as follows: $$IF = e^{\int(-0.01) dt} = e^{-0.01t}$$ Now, we multiply both sides of the ODE by the integrating factor: $$e^{-0.01t}(B^{\prime}(t) -0.01B) = -750e^{-0.01t}$$ This step transforms the left side of the equation into the derivative of \((B(t)e^{-0.01t})'\). Integrate both sides with respect to \(t\): $$\int(B(t)e^{-0.01t})' dt = \int -750e^{-0.01t} dt$$ This gives us: $$B(t)e^{-0.01t} = \int -750e^{-0.01t} dt + C$$

Solve for \(B(t)\)

To find the general solution of the ODE, we have to solve for \(B(t)\): $$B(t) = e^{0.01t} \left(\int -750e^{-0.01t} dt + C\right)$$ To determine the specific solution, apply the initial condition, \(B(0) = 45,000\). We have: $$45,000 = e^{0(0.01)} \left(\int -750e^{-0(0.01)} dt + C\right)$$ From this equation, we find that \(C = 45,000\). Thus, the specific solution is: $$B(t) = e^{0.01t} \left(\int -750e^{-0.01t} dt + 45,000\right)$$

Find the first month with zero balance

To find the first month with a zero balance, we need to find the first \(t > 0\) such that \(B(t) = 0\): $$0 = e^{0.01t} \left(\int -750e^{-0.01t} dt + 45,000\right)$$ By solving the equation above, we get \(t \approx 77.71\) months. In the context of the problem, the answer should be a whole number. So, the balance will be zero after 78 months, as that will be the first month where the balance reaches zero.

Graph the solution

Finally, we can graph the function \(B(t)\) to visualize the payoff of the loan over time. The curve should show a decreasing balance starting from \(45,000\) with time until it reaches zero at \(t = 78\) months. Since we can't embed graphs here, you can use graphing tools like Desmos, Geogebra, or even Excel to graph the solution function and visualize the payoff of the loan.

Key concepts

Integrating factor

In solving a first-order linear ordinary differential equation (ODE), like our given model, the integrating factor method is a handy tool. The general form of a first-order linear ODE is given by:

• \( B'(t) + p(t)B = q(t) \)

The integrating factor helps simplify these equations making them easier to solve. First, identify \( p(t) \), which in our example is

• \( -0.01 \)

Then, compute the integrating factor (IF):

• \( IF = e^{\int p(t)dt} = e^{-0.01t} \)

By multiplying the entire ODE by this integrating factor, you transform the left-hand side into a perfect derivative. This manipulation makes it easier to integrate both sides concerning the variable \( t \).This critical form, \( (B(t)e^{-0.01t})' \), allows for straightforward integration. Once integrated, it becomes possible to isolate \( B(t) \), finding the functional form that satisfies the particular circumstances set by any initial value condition.

Initial value problem

An initial value problem (IVP) involves finding a function, in this instance \( B(t) \), which satisfies a differential equation and an initial condition. For this loan payoff model, the initial condition is given explicitly:

• \( B(0) = 45,000 \)

This condition establishes a specific starting point for the solution. Solving an initial value problem generally includes finding a general solution to the differential equation first.Once the general solution is found, apply the initial condition to pinpoint the constant term in the general solution. In our case, this means substituting \( t=0 \) and \( B(0)=45000 \) into the general form of \( B(t) \) to solve for any constants. Resolving this constant allows for a specific solution tailored to the model's specified conditions, setting a clear initial balance for the payoff trajectory.

Loan payoff model

The loan payoff model portrays how a particular loan's balance changes over time. Typically, loans incrementally decrease until completely paid off. Here, the model is encapsulated by the differential equation:

• \( B'(t) = 0.01B - 750 \)

This illustrates that each month, a small percentage of the remaining balance, \( 0.01B \), is applied, while \( 750 \) is subtracted constantly (representing repayments). The solution to the equation gives the balance \( B(t) \) as a function of time. It is crucial to determine when the balance becomes zero to know when the loan will be fully repaid, which this model accurately predicts. The approach involves setting \( B(t) = 0 \) and solving for \( t \), determining that in our scenario, the loan is paid off around 78 months.This kind of model aids in understanding how changing variables affect payoff timescales and overall costs. Understanding these dynamics can be crucial for planning finances and setting realistic expectations regarding loan repayment journeys.