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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.4.23
Chapter 9, Problem 9.4.23

23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.


B′(t) = 0.005B − 500, B(0) = 50,000

Verified step by step guidance
1
Identify the type of differential equation given: \( B'(t) = 0.005B - 500 \). This is a first-order linear ordinary differential equation.
Rewrite the equation in standard linear form: \( B'(t) - 0.005B = -500 \). This will help us apply the integrating factor method.
Calculate the integrating factor \( \mu(t) \) using the formula \( \mu(t) = e^{\int -0.005 \\ dt} = e^{-0.005t} \).
Multiply both sides of the differential equation by the integrating factor to get \( e^{-0.005t} B'(t) - 0.005 e^{-0.005t} B = -500 e^{-0.005t} \), which simplifies to \( \frac{d}{dt} \left( e^{-0.005t} B \right) = -500 e^{-0.005t} \).
Integrate both sides with respect to \( t \) to find \( e^{-0.005t} B(t) \), then solve for \( B(t) \). Use the initial condition \( B(0) = 50,000 \) to find the constant of integration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Linear Differential Equations

These are differential equations of the form B'(t) + p(t)B = q(t), where the solution involves finding an integrating factor. They model rates of change with linear dependence on the function itself, common in growth and decay problems like loan balances.
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Classifying Differential Equations

Initial Value Problems (IVP)

An IVP specifies the value of the unknown function at a particular point, here B(0) = 50,000. Solving an IVP means finding a unique solution that satisfies both the differential equation and the initial condition, ensuring the model fits the given scenario.
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Initial Value Problems

Interpreting Solutions and Finding Zero Crossing

After solving the differential equation, the solution represents the loan balance over time. Determining when the balance reaches zero involves solving for t when B(t) = 0, which indicates the payoff time. Graphing helps visualize this behavior.
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Finding Area Between Curves that Cross on the Interval Example 2
Related Practice
Textbook Question

Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem

P'(t) = rP (1-P/K), P(0) = P₀

is P(t) = K/((K/P₀ − 1)e⁻ʳᵗ + 1)

Textbook Question

What is a separable first-order differential equation?

Textbook Question

33–42. Solving initial value problems Solve the following initial value problems.

y'(t) = 1 + eᵗ, y(0) = 4

Textbook Question

9–14. Growth rate functions Make a sketch of the population function P (as a function of time) that results from the following growth rate functions. Assume the population at time t = 0 begins at some positive value.


Textbook Question

21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.

u''(x) = 55x⁹ + 36x⁷ - 21x⁵ + 10x⁻³

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Textbook Question

39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form

a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t). 

Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems. 


t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6

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