(a) Explanation The characterization theorem is as follows, Let y 1 and y 2 are linearly independent solutions of the differential equation a y ″ + b y ′ + c y = 0 , that is W ( y 1 , y 2 ) ≠ 0 . Then the general solution of the equation is y = C 1 y 1 + C 2 y 2 for arbitrary constants C 1 and C 2 . To prove the above statement, let, y 1 and y 2 are solutions so that a y 1 ′ ′ ( x ) + b y ′ 1 ( x ) + c y 1 ( x ) = 0 a y 2 ′ ′ ( x ) + b y ′ 2 ( x ) + c y 2 ( x ) = 0 If y = C 1 y 1 + C 2 y 2 , then it can be obtained, a y ″ + b y ′ + c y = a [ C 1 y ″ 1 + C 2 y ″ 2 ] + b [ C 1 y ′ 1 + C 2 y ′ 2 ] + c [ C 1 y 1 + C 2 y 2 ] = C 1 [ a y ″ 1 + b y ′ 1 + c y 1 ] + C 2 [ a y ″ 2 + b y ′ 2 + c y 2 ] = 0 + 0 = 0 Thus, y = C 1 y 1 + C 2 y 2 is also a solution (b) To determine To Discuss: The general solution of a second-order homogeneous equation.

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J

Publisher: Kendall Hunt Publishing

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Chapter 14, Problem 8PE

(a)

To determine

To Discuss:The characteristic equation of ay″+by′+cy=0 .

(b)

To determine

To Discuss:The general solution of a second-order homogeneous equation.

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The characteristic equation of a y ″ + b y ′ + c y = 0 .

Solution Summary: The author explains the characterization theorem of y_1, and the general solution of the equation for arbitrary constants.