To determine The general solution of y ″ + 4 y ′ + ( π 2 + 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 ″ + 4 y ′ + ( π 2 + 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 + 4 λ e λ x + ( π 2 + 4 ) e λ x = 0 e λ x ( λ 2 + 4 λ + ( π 2 + 4 ) ) = 0 λ 2 + 4 λ + ( π 2 + 4 ) = 0 That is, the characteristic equation is λ 2 + 4 λ + ( π 2 + 4 ) = 0 . Compute the roots of the above characteristic equation as follows. λ 2 + 4 λ + ( π 2 + 4 ) = 0 Substitute a = 1, b = 4 and c = ( π 2 + 4 ) in λ = − b ± b 2 − 4 a c 2 a . λ = − 4 ± 4 2 − 4 ( 1 ) ( π 2 + 4 ) 2 ( 1 ) = − 4 ± 4 ⋅ 4 − 4 π 2 − 4 2 = − 4 ± 4 ( 4 − π 2 − 4 ) 2 = − 4 ± 2 − π 2 2 = − 2 + i π or − 2 − i π From the above calculation, it is observed that y ″ + 4 y ′ + ( π 2 + 4 ) y = 0 has complex conjugate roots. Then by case (iii), the basis of y ″ + 4 y ′ + ( π 2 + 4 ) y = 0 is e − 2 x cos π x , e − 2 x sin π x , that is, cos 6 x , sin 6 x . The general solution of y ″ + 4 y ′ + ( π 2 + 4 ) y = 0 is y = e − 2 x ( c 1 cos ( π x ) + c 2 sin ( π x ) ) Now check the correctness of the above solution as follows. Compute the second derivative of y as follows. y ′ = e − 2 x ( − 2 ) ( c 1 cos ( π x ) + c 2 sin ( π x ) ) + e − 2 x ( − π c 1 sin ( π x ) + π c 2 cos ( π x ) ) = − 2 e − 2 x ( c 1 cos ( π x ) + c 2 sin ( π x ) ) + e − 2 x ( − π c 1 sin ( π x ) + π c 2 cos ( π x ) ) y ″ = { − 2 e − 2 x ( − 2 ) ( c 1 cos ( π x ) + c 2 sin ( π x ) ) − 2 e − 2 x ( − π c 1 sin ( π x ) + π c 2 cos ( π x ) ) + + e − 2 x ( − 2 ) ( − π c 1 sin ( π x ) + π c 2 cos ( π x ) ) + e − 2 x ( − π 2 c 1 cos ( π x ) − π 2 c 2 sin ( π x ) ) } = { 4 e −

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ISBN: 9781119480150

Author: Kreyszig

Publisher: WILEY C

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