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Chapter 8: Problem 25

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

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**Q: As a teacher, create a short answer based on the following step by step solution:** If we have the ODE \(B^{\prime}(t) = 0.0075 B - 1500\) and the initial condition \(B(0) = 100,000\), we can solve for B(t) by using the integrating factor method. The solution for B(t) is: \[B(t) = \frac{200,000 - 100,000e^{0.0075t}}{e^{-0.0075t}}\]. If we find when \(B(t) = 0\), we get a value of t 鈮� 92.38. Since t is in months, the loan balance becomes zero after the 93rd month.

Step by step solution

01

Identifying the given ODE and initial condition

The given ODE is: $$B^{\prime}(t)=0.0075 B-1500$$ And the initial condition is: $$B(0)=100,000$$
02

Solving the ODE using the integrating factor method

Rewriting the ODE in the standard form: $$B^{\prime}(t) - 0.0075B = -1500$$ The integrating factor (IF) is given by: $$IF(t)=e^{\int -0.0075 \, dt} = e^{-0.0075t}$$ Multiply the ODE by the integrating factor: $$e^{-0.0075t}B^{\prime}(t) - 0.0075e^{-0.0075t}B = -1500e^{-0.0075t}$$ Now the left side of the equation is the derivative of the product of B(t) and integrating factor: $$\frac{d}{dt}(B(t)e^{-0.0075t}) = -1500e^{-0.0075t}$$ Integrate both sides with respect to t: $$\int \frac{d}{dt}(B(t)e^{-0.0075t}) \, dt = \int -1500e^{-0.0075t} \, dt$$ $$B(t)e^{-0.0075t} = 200,000e^{-0.0075t} + C$$ To find the value of the constant C, apply the initial condition B(0) = 100,000: $$100,000 e^{-0.0075(0)} = 200,000 e^{-0.0075(0)} + C$$ $$C = -100,000$$ The general solution of the ODE is: $$B(t)e^{-0.0075t} = 200,000e^{-0.0075t} - 100,000$$ Now, solve for B(t): $$B(t) = \frac{200,000 - 100,000e^{0.0075t}}{e^{-0.0075t}}$$
03

Finding the first month when the loan balance is zero

Find the time t when B(t) = 0: $$0 = \frac{200,000 - 100,000e^{0.0075t}}{e^{-0.0075t}}$$ Multiplying both sides by \(e^{0.0075t}\), and solving for t, we get: $$200,000 = 100,000e^{0.0075t}$$ Dividing both sides by 100,000, we get: $$2 = e^{0.0075t}$$ Taking the natural log of both sides, we get: $$\ln(2) = 0.0075t$$ Solving for t, we get: $$t = \frac{\ln(2)}{0.0075} \approx 92.38$$ Since t is in months, the first month when the loan balance is zero is after the 93rd month (as it crosses 0 between the 92nd and 93rd months).
04

Graphing the solution

To graph B(t), use its equation: $$B(t) = \frac{200,000 - 100,000e^{0.0075t}}{e^{-0.0075t}}$$ Sketch the graph of B(t) on the interval [0, 100] with t on the x-axis and B(t) on the y-axis. The graph should show that B(t) starts at 100,000 at t = 0 and decreases over time, eventually becoming zero at around t = 92.38, which confirms our previous result that the loan balance is zero after the 93rd month.

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Most popular questions from this chapter

The following models were discussed in Section 1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. A model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object, for \(t \geq 0, g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. Let \(b=0.1 \mathrm{s}^{-1}\). a. Draw the direction field for \(0 \leq t \leq 60,0 \leq y \leq 150\). b. For what initial values \(v(0)=A\) are solutions increasing? Decreasing? c. What is the equilibrium solution?

Suppose an object with an initial temperature of \(T_{0}>0\) is put in surroundings with an ambient temperature of \(A\) where \(A<\frac{T_{0}}{2}\). Let \(t_{1 / 2}\) be the time required for the object to cool to \(\frac{T_{0}}{2}\) a. Show that \(t_{1 / 2}=-\frac{1}{k} \ln \left[\frac{T_{0}-2 A}{2\left(T_{0}-A\right)}\right]\) b. Does \(t_{1 / 2}\) increase or decrease as \(k\) increases? Explain. c. Why is the condition \(A<\frac{T_{0}}{2}\) needed?

Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem $$P^{\prime}(t)=r P\left(1-\frac{P}{K}\right), \quad P(0)=P_{0}$$ $$\text { is } P(t)=\frac{K}{\left(\frac{K}{P_{0}}-1\right) e^{-n}+1}$$

The reaction of certain chemical compounds can be modeled using a differential equation of the form \(y^{\prime}(t)=-k y^{n}(t),\) where \(y(t)\) is the concentration of the compound for \(t \geq 0, k>0\) is a constant that determines the speed of the reaction, and \(n\) is a positive integer called the order of the reaction. Assume that the initial concentration of the compound is \(y(0)=y_{0}>0\). a. Consider a first-order reaction \((n=1)\) and show that the solution of the initial value problem is \(y(t)=y_{0} e^{-k t}\). b. Consider a second-order reaction \((n=2)\) and show that the solution of the initial value problem is \(y(t)=\frac{y_{0}}{y_{0} k t+1}\). c. Let \(y_{0}=1\) and \(k=0.1 .\) Graph the first-order and secondorder solutions found in parts (a) and (b). Compare the two reactions.

Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right),\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-n}},\) where \(C\) is an arbitrary constant. b. Find that value of \(C\) that corresponds to the initial condition \(P(0)=50\). c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\). d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).
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