HW9SOL - HW#9 Chapter 10.1 Prob#5 y ′′ y = x y(0 = y π...

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Unformatted text preview: HW#9 Chapter 10.1 Prob. #5 y ′′ + y = x , y(0) = y( π ) = 0 The homogeneous solution y h ′′ + y h = 0 Assume y h = e rx then, the characteristic equation is r 2 + 1 = 0 Therefore, r = ± ? and y h = Acos(x) + Bsin(x) The particular solution Using the undetermined coefficient method, y p = α x + β y p ′′ = 0 Therefore, α = 1, β = 0 The general solution is y = Acos(x) + Bsin(x) + x Apply the boundary conditions y(0) = A = 0 y( π ) = − A + π = π ≠ Therefore, no solutions exist. Prob. #6. y ′′ + 2y = x , y(0) = y( π ) = 0 The homogeneous solution y h ′′ + 2y h = 0 Assume y h = e rx then, the characteristic equation is r 2 + 2 = 0 Therefore, r = ± ? √ 2 and y h = Acos √ 2 x ¡ + Bsin( √ 2 x) The particular solution Using the undetermined coefficient method, y p = α x + β y p ′′ = 0 Therefore, α = 1/2, β = 0 The general solution is y = Acos √ 2 x ¡ + Bsin √ 2 x ¡ + x 2 Apply the boundary conditions y(0) = A = 0 y( π ) = Bsin( √ 2 π ) + π 2 = 0 Hence, B = − π 2sin( √ 2 π ) y(x) = − π sin √ 2 x...
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This note was uploaded on 10/12/2009 for the course MAE 210792200 taught by Professor Youssef during the Spring '09 term at UCLA.

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HW9SOL - HW#9 Chapter 10.1 Prob#5 y ′′ y = x y(0 = y π...

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