To determine To factor: The ODE P ( D ) = D 2 + 4.00 D + 3.36 I and solve it. 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 P ( D ) = D 2 + 4.00 D + 3.36 I . Factor the ODE P ( D ) = D 2 + 4.00 D + 3.36 I as follows. Let y = e λ x and P ( D ) y = 0 . ( D 2 + 4.00 D + 3.36 I ) y = 0 D 2 y + 4 D y + 3.36 y = 0 ( e λ x ) ′ ′ + 4 ( e λ x ) ′ + 3.36 e λ x = 0 ( e λ x ( λ ) ) ′ + 4 ( e λ x ( λ ) ) + 3

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ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.3, Problem 6P

To determine

To factor: The ODE P(D)=D2+4.00D+3.36I and solve it.

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Chapter 2.3, Problem 5P

Chapter 2.3, Problem 7P

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