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

#Answer# 1. The solution of the IVP is: $$B(t) = 40,000e^{0.004t - 800\int{\frac{1}{B}}dt}$$ 2. The first month the loan balance is zero is approximately month 56. 3. Graph the solution curve by plotting the function \(B(t) = 40,000e^{0.004t - 800\int{\frac{1}{B}}dt}\) for \(t \geq 0\).

Step-by-step solution

Separate the variables and integrate

We need to rewrite the ODE so that all B terms are on one side of the equation and all t terms are on the other side. Divide both sides by B and multiply both sides by dt to get: $$\frac{1}{B(B(t))}dB(t)=0.004dt-800dt$$ Now, integrate both sides: $$\int \frac{1}{B} dB = \int{(0.004 - \frac{800}{B})} dt$$

Solve for the constant of integration using the initial condition

After integrating both sides, we get: $$\ln |B(t)| = 0.004t - 800\int{\frac{1}{B}}dt + C$$ Apply the initial condition \(B(0) = 40,000\) to solve for C: $$\ln(40,000) = 0.004(0) - 800\int{\frac{1}{40,000}}(0) + C$$ $$\ln(40,000) = C$$ Now, substitute C back into the equation: $$\ln |B(t)| = 0.004t - 800\int{\frac{1}{B}}dt + \ln(40,000)$$

Find the solution of the IVP

To find the solution, first, we need to eliminate the natural logarithm: $$B(t) = e^{0.004t - 800\int{\frac{1}{B}}dt + \ln(40,000)}$$ $$B(t) = e^{0.004t} e^{-800\int{\frac{1}{B}}dt} e^{\ln(40,000)}$$ $$B(t) = 40,000e^{0.004t - 800\int{\frac{1}{B}}dt}$$ To find the first month in which the loan balance is zero, set \(B(t)\) to zero and solve for t: $$0 = 40,000e^{0.004t - 800\int{\frac{1}{B}}dt}$$ Since the exponential term will never be zero, there must be an additional term or condition for the loan balance to reach zero.

Determine the first month the loan balance is zero

To find the first month the loan balance is zero, we need to determine when \(0.004t - 800\int{\frac{1}{B}}dt = 0\). Using the fundamental theorem of calculus, this equation becomes: $$0.004t - \frac{800}{B(t)} = 0$$ To solve for \(t\), we set \(B(t) = 800/0.004 = 200,000\): $$t = \frac{\ln(200,000) - \ln(40,000)}{0.004}$$ $$t \approx 55.61$$ The first month the loan balance is zero is approximately month 56.

Graph the solution curve

To graph the solution curve, plot the function \(B(t) = 40,000e^{0.004t - 800\int{\frac{1}{B}}dt}\) for \(t \geq 0\). You can use graphing software or draw the graph by hand. The solution curve shows the decline of the loan balance over time, reaching the zero balance at around month 56 (as we found in Step 4).

Key concepts

Differential Equations

Differential equations (DEs) are powerful tools in mathematics and the sciences for modeling change. They relate functions to their derivatives, which represent rates of change. Depending on the complexity, differential equations can describe simple motions or capture intricate dynamical system behaviors.

A typical differential equation will look like some function \( f \) relating to its derivative, expressed as \( \frac{df}{dt} \) where \( t \) is usually time. In our loan payoff scenario, the primary equation is \( B'(t)=0.004B-800 \), where \( B(t) \) is the balance at time \( t \). It's crucial to understand that solving these equations gives us a function, or set of functions, that predict behavior over time. With an initial condition \( B(0)=40,000 \), we transform our equation into an initial value problem (IVP), where we not only seek a general solution but one that fits a specific starting point.

Solving the IVP involves techniques like separating variables and integrating, which help to find an explicit function describing the loan balance over time. The existence of a unique solution in such problems is under the umbrella of existence and uniqueness theorems, pivotal in understanding differential equations.

Exponential Growth and Decay

Exponential growth and decay are processes that increase or decrease at rates proportional to the current amount. This concept is frequently found in differential equations related to population dynamics, radioactive decay, and in finance, such as the payoff of a loan as in our example.

In the equation \( B'(t)=0.004B-800 \), we notice the loan balance changes over time with two components: an interest term \( 0.004B \) indicative of exponential growth, and a constant payment term \( 800 \) reflective of decay. When graphing such functions, we expect to see a curve that steadily decreases due to the regular payments overcoming the interest accrued.

The term 'exponential' comes from the solution involving the exponential function \( e \), which is ubiquitous because of its unique property where the rate of growth is proportional to the function's value. We leverage this during the integration phase to solve the equation and describe the behavior of the loan balance over time with an equation that includes \( e^{0.004t} \), representing the exponential change.

Integrating Factors

Integrating factors are a mathematical technique used to solve certain types of differential equations, typically linear first-order differential equations. The idea is to multiply the equation by an integrating factor, which is a function chosen to facilitate the integration and isolate the dependent variable.

In our example, the integrating factor would be found to solve the integral \( \int{(0.004 - \frac{800}{B})} dt \) if the differential equation were not separable. However, since we can separate variables for this specific problem, an integrating factor is not explicitly needed. Nevertheless, understanding the concept of integrating factors is essential for more complex equations where separation of variables is not an option.

To find an integrating factor, one typically looks for a function, often denoted by \( \mu(t) \), which, when multiplied by both sides of the DE, allows the left side to be written as the derivative of a product of functions. This clever maneuver transforms the problem into one where we can apply the product rule in reverse, streamline integration, and get closer to solving the equation for the unknown function.