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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 9.4.47
Chapter 9, Problem 9.4.47

45–48. General first-order linear equations Consider the general first-order linear equation y'(t)+a(t)y(t)=f(t). This equation can be solved, in principle, by defining the integrating factor p(t)=exp(∫a(t)dt). Here is how the integrating factor works. Multiply both sides of the equation by p (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes


p(t)(y′(t) + a(t)y(t)) = d/dt(p(t)y(t)) = p(t)f(t).


Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor.


y′(t) + (2t)/(t² + 1)y(t) = 1 + 3t², y(1) = 4

Verified step by step guidance
1
Identify the given first-order linear differential equation: \(y'(t) + \frac{2t}{t^2 + 1} y(t) = 1 + 3t^2\), with the initial condition \(y(1) = 4\).
Determine the integrating factor \(p(t)\) using the formula \(p(t) = \exp\left(\int a(t) \, dt\right)\), where \(a(t) = \frac{2t}{t^2 + 1}\). So, compute \(p(t) = \exp\left(\int \frac{2t}{t^2 + 1} \, dt\right)\).
Evaluate the integral inside the exponent to find \(p(t)\). Notice that the integral involves a rational function that can be simplified by substitution.
Multiply both sides of the original differential equation by the integrating factor \(p(t)\) to rewrite the left side as the derivative of the product \(p(t) y(t)\): \(\frac{d}{dt} \left(p(t) y(t)\right) = p(t) (1 + 3t^2)\).
Integrate both sides with respect to \(t\) to get \(p(t) y(t) = \int p(t) (1 + 3t^2) \, dt + C\). Then solve for \(y(t)\) by dividing both sides by \(p(t)\), and use the initial condition \(y(1) = 4\) to find the constant \(C\).

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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

A first-order linear differential equation has the form y' + a(t)y = f(t), where a(t) and f(t) are functions of t. Such equations describe many physical and mathematical processes and can be solved systematically using integrating factors. Understanding the structure of these equations is essential for applying solution techniques.
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Classifying Differential Equations

Integrating Factor Method

The integrating factor is a function p(t) = exp(∫a(t) dt) used to simplify first-order linear equations. Multiplying the entire differential equation by p(t) transforms the left side into the derivative of p(t)y(t), making it easier to integrate both sides. This method converts the equation into an exact derivative form, facilitating straightforward integration.
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Euler's Method

Initial Value Problems (IVP)

An initial value problem specifies the value of the solution y(t) at a particular point t = t₀, such as y(1) = 4. Solving an IVP involves finding the general solution of the differential equation and then using the initial condition to determine the unique constant of integration, ensuring a specific solution that fits the given initial data.
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Related Practice
Textbook Question

25–28. Two steps of Euler’s method For the following initial value problems, compute the first two approximations u1 and u2 given by Euler’s method using the given time step.


y′(t) = 2−y, y(0) = 1; Δt = 0.1

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Textbook Question

12–16. Sketching direction fields Use the window [-2, 2] x [-2, 2] to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. A detailed direction field is not needed.

y(x) = sin y, y(−2) = 1/2

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. 


(t² + 1)y′(t) + 2ty = 3t², y(2) = 8

Textbook Question

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) = 0.004B − 800, B(0) = 40,000

Textbook Question

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

y'(t) = t lnt + 1

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Textbook Question

5–16. Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable.

u'(x) = e²ˣ⁻ᵘ