Textbook Question
17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = y³sin t, y(0) = 1
Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.4.6
5–10. First-order linear equations Find the general solution of the following equations.
y'(x) = −y + 2
Verified step by step guidance1
Rewrite the differential equation in the standard linear form: \(y'(x) + y = 2\).
Identify the integrating factor \(\mu(x)\), which is given by \(\mu(x) = e^{\int 1 \, dx} = e^{x}\).
Multiply both sides of the equation by the integrating factor to get: \(e^{x} y' + e^{x} y = 2 e^{x}\).
Recognize that the left side is the derivative of the product \(e^{x} y\), so write it as \(\frac{d}{dx} (e^{x} y) = 2 e^{x}\).
Integrate both sides with respect to \(x\): \(\int \frac{d}{dx} (e^{x} y) \, dx = \int 2 e^{x} \, dx\), then solve for \(y\) by dividing by \(e^{x}\) and adding the constant of integration.
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1mKey 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' + p(x)y = q(x). It involves the first derivative of the unknown function and can be solved using integrating factors or other standard methods. Recognizing this form is essential to apply the correct solution technique.
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Classifying Differential Equations
Integrating Factor Method
The integrating factor method involves multiplying the entire differential equation by a specially chosen function, usually e^(∫p(x)dx), to rewrite the left side as a derivative of a product. This simplifies solving the equation by allowing direct integration.
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General Solution of Differential Equations
The general solution includes all possible solutions of a differential equation and typically contains an arbitrary constant. It combines the homogeneous solution (solving y' + p(x)y = 0) and a particular solution to the nonhomogeneous equation.
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Related Practice
Textbook Question
The general solution of a first-order linear differential equation is y(t) = Ce⁻¹⁰ᵗ − 13. What solution satisfies the initial condition y(0) = 4?
Textbook Question
21–24. Logistic equations Consider the following logistic equations. In each case, sketch the direction field, draw the solution curve for each initial condition, and find the equilibrium solutions. A detailed direction field is not needed. Assume t ≥ 0 and tP ≥ 0.
P′(t) = 0.05P(1−P/800); P(0) = 100, P(0) = 400, P(0) = 700
Textbook Question
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
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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'(t) = 4−y, y(0) = −1
Textbook Question
33–42. Solving initial value problems Solve the following initial value problems.
y'(x) = 4 sec² 2x, y(0) = 8