Explanation Formula used: Cosine formula: cos ( A + B ) = cos ( A ) cos ( B ) − sin ( A ) sin ( B ) cos ( A − B ) = cos ( A ) cos ( B ) + sin ( A ) sin ( B ) Calculation: Consider the differential equation L [ y ] = [ D 2 − 2 λ D + ( λ 2 + μ 2 ) ] y = g ( t ) with initial conditions y ( t 0 ) = 0 and y ′ ( t 0 ) = 0 . Given that the roots of characteristic equation are λ ± i μ . This implies that the complementary solution is y c ( t ) = A e λ t cos μ t + B e λ t sin μ t . Take y 1 ( t ) = e λ t cos μ t and y 2 ( t ) = e λ t sin μ t . Compute the Wronskian W ( t ) as follows. W ( t ) = | y 1 y 2 y 1 ′ y 2 ′ | = | e λ t cos μ t e λ t sin μ t λ e λ t cos μ t − μ e λ t sin t λ e λ t sin μ t + μ e λ t cos μ t | = λ e 2 λ t sin μ t cos μ t + μ e 2 λ t cos 2 μ t − λ e 2 λ t sin μ t cos μ t + μ e 2 λ t sin 2 t = μ e 2 λ t From problem 22, it is obtained that the particular solution of L [ y ] = y ″ + p ( t ) y ′ + q ( t ) y = g ( t ) is Y ( t ) = ∫ t 0 t y 2 ( t ) y 1 ( s ) − y 1 ( t ) y 2 ( s ) y 1 ( s ) y 2 ′ ( s ) − y 2 ( s ) y 1 ′ ( s ) g ( s ) d s

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ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3.6, Problem 25P

To determine

The solution of the initial value problem L[y]=[D2−2λD+(λ2+μ2)]y=g(t) with initial conditions y(t0)=0 and y′(t0)=0.

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Chapter 3.6, Problem 24P

Chapter 3.6, Problem 26P

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Chapter 3 Solutions

Elementary Differential Equations

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The solution of the initial value problem L [ y ] = [ D 2 − 2 λ D + ( λ 2 + μ 2 ) ] y = g ( t ) with initial conditions y ( t 0 ) = 0 and y ′ ( t 0 ) = 0 .

The solution of the initial value problem L[y]=[D2−2λD+(λ2+μ2)]y=g(t) with initial conditions y(t0)=0 and y′(t0)=0.

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Chapter 3 Solutions