To determine The general solution of y ″ + 2 k 2 y ′ + k 4 y = 0 ; Check the answer by substitution. Explanation Result used: Case (i): If the auxiliary equation has distinct real roots λ 1 , and λ 2 of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e λ 1 x , e λ 2 x and the general solution of y ″ + a y ′ + b y = 0 is y = c 1 e λ 1 x + c 2 e λ 2 x . Case (ii): If the auxiliary equation has real double root λ = − a of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e − a x , x e − a x and the general solution of y ″ + a y ′ + b y = 0 is y = ( c 1 + c 2 ) e − a x . Case (iii): If the auxiliary equation has complex conjugate root λ 1 = a + i ω , λ 2 = a − i ω of y = e λ x , then the basis of y ″ + a y ′ + b y = 0 is e a x cos ω x , e a x sin ω x and the general solution of y ″ + a y ′ + b y = 0 is y = e − a x ( A cos ω x + B sin ω x ) . Calculation: The given ODE is y ″ + 2 k 2 y ′ + k 4 y = 0 (1) Compute the auxiliary equation of the above ODE as follows. Let y = e λ x . That implies, the first and second derivative of y is y ′ = λ e λ x and y ″ = λ 2 e λ x , respectively. Substitute y ″ = λ 2 e λ x and y = e λ x in (1) and simplify further. λ 2 e λ x + 2 k 2 λ e λ x + k 4 e λ x = 0 e λ x ( λ 2 + 2 k 2 λ + k 4 ) = 0 λ 2 + 2 k 2 λ + k 4 = 0 That is, the characteristic equation is λ 2 + 2 k 2 λ + k 4 = 0 . Compute the roots of the above characteristic equation as follows. 9 λ 2 − 30 λ + 25 = 0 Substitute a = 1, b = 2 k 2 and c = k 4 in λ = − b ± b 2 − 4 a c 2 a . λ = − 2 k 2 ± ( 2 k 2 ) 2 − 4 ( 1 ) ( k 4 ) 2 ( 1 ) = − 2 k 2 ± 4 k 4 − 4 k 4 2 = − 2 k 2 2 = − k 2 From the above calculation, it is observed that y ″ + 2 k 2 y ′ + k 4 y = 0 has real double root. Then by case (iii), the basis of y ″ + 2 k 2 y ′ + k 4 y = 0 is e − k 2 x , x e − k 2 x . The general solution of y ″ + 2 k 2 y ′ + k 4 y = 0 is y = e − k 2 x ( c 1 + c 2 x ) . Now check the correctness of the above solution as follows. Compute the second derivative of y as follows

Custom Kreyszig: Advanced Engineering Mathematics

10th Edition

ISBN: 9781119166856

Author: Kreyszig

Publisher: JOHN WILEY+SONS INC.CUSTOM

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The general solution of y″+2k2y′+k4y=0; Check the answer by substitution.

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