Explanation Given: The given initial boundary value problem is: y ″ − 4 y ′ + 13 y = 0 ; y ( 0 ) = 0 , y ( π ) = 0 Procedure used: Solution for homogeneous equation: Consider the equation, a y ″ + b y ′ + c y = 0 . First find the roots of the auxiliary equation: a r 2 + b r + c = 0 If the roots are conjugate complex α ± β i , then the general solution form is: y ( x ) = c 1 e α x cos β x + c 2 e α x sin β x After that, use the initial condition to find the value of the constants in the general solution. Quadratic formula for finding the roots: − b ± b 2 − 4 a c 2 a Here, a, b and c are coefficients of the quadratic equation. Calculation: Consider the given initial boundary value problem: y ″ − 4 y ′ + 13 y = 0 ; y ( 0 ) = 0 , y ( π ) = 0 The auxiliary equation is: r 2 − 4 r + 13 = 0 Find the roots of the equation by using the quadratic formula: r = − ( − 4 ) ± ( − 4 ) 2 − 4 × 1 × 13 2 = 4 ± 16 − 52 2 = 4 ± − 36 2 = 2

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 11.2, Problem 3E

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The solution for the given initial boundary value problem y″−4y′+13y=0; y(0)=0, y(π)=0.

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Chapter 11.2, Problem 2E

Chapter 11.2, Problem 4E

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Chapter 11 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - Prob. 3E Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - Prob. 7E Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - Prob. 9E Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...

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The solution for the given initial boundary value problem y ″ − 4 y ′ + 13 y = 0 ; y ( 0 ) = 0 , y ( π ) = 0 .

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The solution for the given initial boundary value problem y″−4y′+13y=0; y(0)=0, y(π)=0.

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