Ch. 11 - Power Series

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 11 - Power Series

Problem 11.R.63

Chapter 11, Problem 11.R.63

A differential equation Find a power series solution of the differential equation y'(x) - 4y + 12 = 0, subject to the condition y(0) = 4. Identify the solution in terms of known functions.

Verified step by step guidance

Rewrite the differential equation in standard form: \(y'(x) = 4y - 12\).

Assume a power series solution of the form \(y(x) = \sum_{n=0}^{\infty} a_n x^n\) and find its derivative \(y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\).

Substitute the series expressions for \(y(x)\) and \(y'(x)\) into the differential equation to get \(\sum_{n=1}^{\infty} n a_n x^{n-1} = 4 \sum_{n=0}^{\infty} a_n x^n - 12\).

Align powers of \(x\) on both sides by shifting indices as needed, then equate coefficients of like powers of \(x\) to find a recurrence relation for the coefficients \(a_n\).

Use the initial condition \(y(0) = 4\) to find \(a_0\), then solve the recurrence relation to express \(y(x)\) as a power series and recognize it as a known function (likely involving exponentials).

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Key Concepts

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

Power Series Solutions of Differential Equations

A power series solution expresses the unknown function as an infinite sum of powers of the variable, typically centered at a point like x=0. This method is useful when standard solution techniques are difficult, allowing the differential equation to be solved by determining the coefficients of the series.

Initial Conditions and Their Role

Initial conditions specify the value of the solution or its derivatives at a particular point, enabling the determination of arbitrary constants in the general solution. For example, y(0) = 4 fixes the constant term in the power series, ensuring the solution fits the given problem.

Solving First-Order Linear Differential Equations

First-order linear differential equations have the form y' + p(x)y = q(x). They can be solved using integrating factors or by recognizing standard solution forms. Identifying the solution in terms of known functions often involves rewriting the equation and integrating accordingly.

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Related Practice

Textbook Question

Radius and interval of convergence Use the Ratio Test or the Root Test to determine the radius of convergence of the following power series. Test the endpoints to determine the interval of convergence, when appropriate.

∞

Σ (x - 1)ᵏ/(k5ᵏ)

k = 1

Textbook Question

Limits by power series Use Taylor series to evaluate the following limits.

lim ₙ → 0 (x²/2 - 1 + cos x)/x⁴

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Textbook Question

∞

Σ x⁴ᵏ/k²

k = 1

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Textbook Question

Approximating real numbers Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. There is more than one way to choose the center of the series.

sinh (-1)

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Textbook Question

Approximating ln 2 Consider the following three ways to approximate

ln 2.

d. At what value of x should the series in part (c) be evaluated to approximate ln 2? Write the resulting infinite series for ln 2.

Textbook Question

Find the remainder term Rₙ(x) for the Taylor series centered at 0 for the following functions. Find an upper bound for the magnitude of the remainder on the given interval for the given value of n. (The bound is not unique.)

ƒ(x) = eˣ; bound R₃(x), for |x| < 1