Explanation The Radius of convergence for any series basically tells about the domain on which that series is convergent. Given: P ( x ) = ∑ n = 0 ∞ a n x n satisfying the differential equation y ' = − y with initial condition y ( 0 ) = 1 . Theorem 1: If y is a differentiable function satisfying the differential equation y ' = k y then y ( t ) = P 0 e k t ,Where P 0 is the initial value P 0 = y ( 0 ) . Calculation: We have y = P ( x ) = ∑ n = 0 ∞ a n x n = a 0 + a 1 x + a 2 x 2 + ... y ' = P ' ( x ) = ∑ n = 0 ∞ n a n x n − 1 = a 1 + 2 a 2 x + 3 a 3 x 2 + ... Now, the differential equation is: y ' = − y y ' + y = 0 ... ( 1 ) Now, substitute the values of y and y’ into the equation (1) and we get: ∑ n = 0 ∞ a n x n + ∑ n = 0 ∞ n a n x n − 1 = 0 ∑ n = 0 ∞ a n x n + ∑ n = 1 ∞ n a n x n = 0 We see that the equation (1) will satisfy if: a 0 = − a 1 , a 1 = − 2 a 2 and a 2 = − 3 a 3 In general, we can say that

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 11.6, Problem 57E

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A power series $P (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ satisfying the differential equation $y' = - y$ with initial condition $y (0) = 1$ . Then use Theorem 1 to conclude that $P (x) = e^{- x}$ .

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Chapter 11.6, Problem 56E

Chapter 11.6, Problem 58E

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Chapter 11 Solutions

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Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E

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A power series P ( x ) = ∑ n = 0 ∞ a n x n satisfying the differential equation y ' = − y with initial condition y ( 0 ) = 1 . Then use Theorem 1 to conclude that P ( x ) = e − x .

Solution Summary: Theorem 1: The Radius of convergence for any series basically tells about the domain on which that series is convergent.

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To calculate:

A power series $P (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ satisfying the differential equation $y' = - y$ with initial condition $y (0) = 1$ . Then use Theorem 1 to conclude that $P (x) = e^{- x}$ .

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Chapter 11 Solutions

A power series P ( x ) = ∑ n = 0 ∞ a n x n satisfying the differential equation y ' = − y with initial condition y ( 0 ) = 1 . Then use Theorem 1 to conclude that P ( x ) = e − x .

Solution Summary: Theorem 1: The Radius of convergence for any series basically tells about the domain on which that series is convergent.

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To calculate: A power series P(x)=∑n=0∞anxn satisfying the differential equation y'=−y with initial condition y(0)=1. Then use Theorem 1 to conclude that P(x)=e−x.

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Chapter 11 Solutions

To calculate:

A power series $P (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ satisfying the differential equation $y' = - y$ with initial condition $y (0) = 1$ . Then use Theorem 1 to conclude that $P (x) = e^{- x}$ .