Explanation Given information: The differential equation, y ″ − y ′ − 12 y = 0 General solution of the differential equation is y = C 1 e 4 x + C 2 e − 3 x Initial conditions are y = 5 and y ′ = 6 when x = 0 Consider the differential equations, y ″ − y ′ − 12 y = 0 …… (1) Consider the general solution as, y = C 1 e 4 x + C 2 e − 3 x …… (2) Differentiate equation (2) with respect to x , y ′ = 4 C 1 e 4 x − 3 C 2 e − 3 x …… (3) Again differentiate above equation, y ″ = 16 C 1 e 4 x + 9 C 2 e − 3 x Now substitute the values of y ′ and y ″ in equation (1), ( 16 C 1 e 4 x + 9 C 2 e − 3 x ) − ( 4 C 1 e 4 x − 3 C 2 e − 3 x ) − 12 ( C 1 e 4 x + C 2 e − 3 x ) = ? 0 16 C 1 e 4 x + 9 C 2 e − 3 x − 4 C 1 e 4 x + 3 C 2 e − 3 x − 12 C 1 e 4 x − 12 C 2 e − 3 x = ? 0 16 C 1 e 4 x + 9 C 2 e − 3 x − 16 C 1 e 4 x − 9 C 2 e − 3 x = ? 0 0 = ? 0 This is true. Hence, y = C 1 e 4 x + C 2 e − 3 x is the solution of the differential equation y ″ − y ′ − 12 y = 0

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ISBN: 8220101426222

Author: Larson

Publisher: CENGAGE L

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Chapter 11.1, Problem 28E

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Whether y=C1e4x+C2e−3x is the general solution of the differential equation y″−y′−12y=0 and then find the particular solution of the differential equation when the initial conditions are y=5 and y′=6 when x=0.

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Chapter 11.1, Problem 27E

Chapter 11.1, Problem 29E

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Whether y = C 1 e 4 x + C 2 e − 3 x is the general solution of the differential equation y ″ − y ′ − 12 y = 0 and then find the particular solution of the differential equation when the initial conditions are y = 5 and y ′ = 6 when x = 0 .

Solution Summary: The author explains that the differential equation y=C_1e4x+

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Whether y=C1e4x+C2e−3x is the general solution of the differential equation y″−y′−12y=0 and then find the particular solution of the differential equation when the initial conditions are y=5 and y′=6 when x=0.

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