Explanation Given: Initial-conditions: y ( 0 ) = y 0 and y ′ ( 0 ) = y 1 Three differential equations are: x 2 y ″ = 0 , x 2 y ″ − 2 x y ′ + 2 y = 0 , x 2 y ″ − 4 x y ′ + 6 y = 0. Calculation: Consider the Cauchy-Euler’s DE x 2 y ″ = 0 . Substitute x m into the above DE, x 2 m ( m − 1 ) x m − 2 = 0 x m m ( m − 1 ) = 0 For any m , x m ≠ 0 . m ( m − 1 ) = 0 m = 0 , m = 1 Therefore, the roots of auxiliary equation of DE x 2 y ″ = 0 are m 1 = 0 and m 2 = 1 . Thus, the general solution is y ( x ) = c 1 + c 2 x . Substitute the value y ( 0 ) = y 0 in y ( x ) = c 1 + c 2 x , y ( 0 ) = c 1 + c 2 ( 0 ) y 0 = c 1 Differentiate y ( x ) = c 1 + c 2 x with respect to x . y ′ ( x ) = c 2 Substitute the value y ′ ( 0 ) = y 1 in the above equation, y ′ ( 0 ) = c 2 c 2 = y 1 Substitute the values for c 1 and c 2 in y ( x ) = c 1 + c 2 x , y ( x ) = y 0 + y 1 x Thus, the general solution is y ( x ) = y 0 + y 1 x and the initial conditions are satisfied for all y 0 , y 1 ∈ R . Consider the Cauchy-Euler’s DE x 2 y ″ − 2 x y ′ + 2 y = 0 . Substitute x m into the above DE, x 2 y ″ − 2 x y ′ + 2 y = 0 x 2 m ( m − 1 ) x m − 2 − 2 x m x m − 1 + 2 x m = 0 x m ( m 2 − m − 2 m + 2 ) = 0 x m ( m 2 − 3 m + 2 ) = 0 For any m , x m ≠ 0 . m 2 − 3 m + 2 = 0 ( m − 1 ) ( m − 2 ) = 0 m = 1 , 2 The roots of auxiliary equation of DE x 2 y ″ − 2 x y ′ + 2 y = 0 are m 1 = 1 and m 2 = 2

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 3 Solutions

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Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

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