Explanation Given: The function with period 2 is f ( t ) = { 1 − t , 0 ≤ t < 1 − 1 + t , 1 ≤ t < 2 . Calculation: Given initial value problem y ″ + ω 2 y = f ( t ) with initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = 0 . From Exercise 16 of section 10.2, the general solution of the function f ( x ) = { 1 − x , 0 ≤ x < 1 1 − x , 1 ≤ x < 2 with period of 2 is is f ( x ) = 1 2 + 4 π 2 ∑ n = 1 ∞ cos ( ( 2 n − 1 ) π x ) ( 2 n − 1 ) 2 . The homogenous solution for the above general solution can be written as y c ( t ) = c 1 cos ω t + c 2 sin ω t . Compute the particular integral of the homogeneous differential equation y ″ + ω 2 y = f ( t ) as shown below. Consider the solution has a Fourier series transform y p ( t ) = c 0 + 4 π 2 ∑ n = 1 ∞ c n cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 . Compute the first derivative for y = c 0 + ∑ n = 1 ∞ c n cos ( n π t ) . y ′ ( t ) = ( c 0 + ∑ n = 1 ∞ c n cos ( n π t ) ) ′ = 0 − ∑ n = 1 ∞ n π c n sin ( n π t ) = − ∑ n = 1 ∞ n π c n sin ( n π t ) Compute the second derivative of the above solution. y ″ ( t ) = ( c 0 + ∑ n = 1 ∞ c n cos ( n π t ) ) ′ ′ = ( − ∑ n = 1 ∞ n π c n sin ( n π t ) ) ′ = − ∑ n = 1 ∞ n 2 π 2 c n cos ( n π t ) = − ∑ n = 1 ∞ ( n − 1 ) 2 π 2 c n cos ( n π t ) Now substitute the second derivative and y in the given differential equation y ″ + ω 2 y = f ( t ) . y ″ + ω 2 y = f ( t ) − ∑ n = 1 ∞ ( n − 1 ) 2 π 2 c n cos ( n π t ) + ω 2 ( c 0 + ∑ n = 1 ∞ c n cos ( n t ) ) = 1 2 + ∑ n = 1 ∞ a n cos ( n π t ) − ∑ n = 1 ∞ ( n − 1 ) 2 π 2 c n cos ( n π t ) + ω 2 c 0 + ω 2 ∑ n = 1 ∞ c n cos ( n t ) = 1 2 + ∑ n = 1 ∞ a n cos ( n π t ) ∑ n = 1 ∞ c n cos ( n π t ) ( ω 2 − ( n − 1 ) 2 π 2 ) + ω 2 c 0 = 1 2 + ∑ n = 1 ∞ a n cos ( n π t ) ⋯ ⋯ ( 1 ) Equate the coefficient of constant terms in the equation (1). ω 2 c 0 = 1 2 c 0 = 1 2 ω 2 Equate the coefficient of ∑ n = 1 ∞ cos ( n π t ) in the equation (1). ( ω 2 − n 2 π 2 ) c n = a n c n = a n ( ω 2 − ( n − 1 ) 2 π 2 ) Thus, c 0 = 1 2 ω 2 and c n = a n ( ω 2 − ( n − 1 ) 2 π 2 ) . Substitute the values c 0 = 1 2 ω 2 and c n = a n ( ω 2 − ( n − 1 ) 2 π 2 ) in y p ( t ) = c 0 + 4 π 2 ∑ n = 1 ∞ c n cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 . y p ( t ) = c 0 + 4 π 2 ∑ n = 1 ∞ c n cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 = 1 2 ω 2 + 4 π 2 ∑ n = 1 ∞ a n cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 ( ω 2 − ( n − 1 ) 2 π 2 ) The general solution of y ″ + ω 2 y = f ( t ) is given by y = y h + y p . Hence, y = c 1 cos ω t + c 2 sin ω t + 1 2 ω 2 + 4 π 2 ∑ n = 1 ∞ a n cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 ( ω 2 − ( n − 1 ) 2 π 2 ) is the formal general solution of the differential equation y ″ + ω 2 y = f ( t ) . Use the given initial conditions y ( 0 ) = 1 , y ′ ( 0 ) = 0 to evaluate c 1 and c 2 . Substitute the values t = 0 and y ( 0 ) = 1 in the general solution y = c 1 cos ω t + c 2 sin ω t + 1 2 ω 2 + 4 π 2 ∑ n = 1 ∞ a n cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 ( ω 2 − ( n − 1 ) 2 π 2 ) as follows. y ( 0 ) = 1 c 1 cos ω ( 0 ) + c 2 sin ω ( 0 ) + 1 2 ω 2 + 4 π 2 ∑ n = 1 ∞ cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 ( ω 2 − n 2 π 2 ) = 1 c 1 = 1 − 1 2 ω 2 − 4 π 2 ∑ n = 1 ∞ cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 ( ω 2 − ( n − 1 ) 2 π 2 ) Compute the derivative of y = c 1 cos ω t + c 2 sin ω t + 1 2 ω 2 + 4 π 2 ∑ n = 1 ∞ a n cos ( ( 2 n − 1 ) π t ) ( 2 n − 1 ) 2 ( ω 2 − ( n − 1 ) 2 π 2 )

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Chapter 10.3, Problem 16P

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The formal solution of an initial value problem y″+ω2y=f(t) with initial conditions y(0)=1 and y′(0)=0.

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The formal solution of an initial value problem y ″ + ω 2 y = f ( t ) with initial conditions y ( 0 ) = 1 and y ′ ( 0 ) = 0 .

The formal solution of an initial value problem y″+ω2y=f(t) with initial conditions y(0)=1 and y′(0)=0.

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