Textbook Question
Evaluate the integrals in Exercises 77–90.
79. ∫(from -1 to 0)6dt/√(3-2t-t²)
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.4.5
In Exercises 5–8, show that each function is a solution of the given initial value problem.
5. Differential Equation: 2y + y' = 4x + 2
Initial condition: y(-1) = e² - 2
Solution candidate: y = e^(-2x) + 2x
Verified step by step guidance1
Identify the given differential equation: \(2y + y' = 4x + 2\) and the initial condition \(y(-1) = e^{2} - 2\).
Calculate the derivative of the candidate solution \(y = e^{-2x} + 2x\). Use the chain rule for the exponential term: \(y' = \frac{d}{dx}(e^{-2x}) + \frac{d}{dx}(2x) = -2e^{-2x} + 2\).
Substitute the candidate solution \(y\) and its derivative \(y'\) into the left side of the differential equation: \(2y + y' = 2(e^{-2x} + 2x) + (-2e^{-2x} + 2)\).
Simplify the expression from the substitution to verify if it equals the right side of the differential equation \(4x + 2\). This confirms whether the candidate function satisfies the differential equation.
Check the initial condition by substituting \(x = -1\) into the candidate solution \(y = e^{-2(-1)} + 2(-1)\) and verify if it equals \(e^{2} - 2\). This confirms the solution meets the initial value condition.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Initial Value Problem (IVP)
An initial value problem consists of a differential equation paired with a specific condition that the solution must satisfy at a given point. This condition, called the initial condition, ensures a unique solution by specifying the value of the function at a particular input, such as y(-1) = e² - 2.
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Initial Value Problems
Verification of a Solution to a Differential Equation
To verify a candidate solution, substitute it and its derivative into the differential equation to check if the equation holds true for all x in the domain. This process confirms whether the proposed function satisfies the relationship defined by the differential equation.
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Differentiation of Exponential and Polynomial Functions
Differentiation rules for exponential functions like e^(-2x) and polynomial terms such as 2x are essential to find y'. The derivative of e^(-2x) is -2e^(-2x), and the derivative of 2x is 2, which are used to substitute into the differential equation for verification.
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Related Practice
Textbook Question
Solve the initial value problems in Exercises 115–120.
117. dy/dx = 1/(x√(x² - 1)), x > 1; y(2) = π
Textbook Question
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
24. lim (x → π/2) (ln(csc x)) / (x - (π/2))²
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Textbook Question
Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76.
Try it—you just keep on cycling. Find the limits some other way.
73. lim (x → ∞) (2^x - 3^x) / (3^x + 4^x)
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Textbook Question
Evaluate the integrals in Exercises 33–54.
∫ (e^(√r) / √r) dr
Textbook Question
Show that the graph of the inverse of f(x)=mx+b, where m and b are constants and m≠0, is a line with slope 1/m and y-intercept -b/m.