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Chem Differential Eq HW Solutions Fall 2011 127

# Chem Differential Eq HW Solutions Fall 2011 127 - Section...

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Section 7.7 Problems Involving Semi-Infinite Intervals 127 or d dt u s ( w,t ) + c 2 ω 2 u s ( w,t ) = c 2 2 π wT 0 . Taking the Fourier sine transform of the boundary condition u ( x, 0) = 0 for x> 0, we get u s ( w, 0) = 0. (b) A particular solution of the differential equation can be guessed easily: u s ( w,t ) = 2 π T 0 w . The general solution of the homogeneous differential equation: d dt u s ( w,t ) + c 2 ω 2 u s ( w,t ) = 0 is u s ( w,t ) = A ( w ) e - c 2 w 2 t . So the general solution of the nonhomogeneous differ- ential equation is u s ( w,t ) = A ( w ) e - c 2 w 2 t 2 π T 0 w . Using u s ( w, 0) = A ( w ) 2 π T 0 w = 0, we find A ( w ) = - 2 π T 0 w . So u s ( wt ) = 2 π T 0 w - 2 π T 0 w e - c 2 w 2 t . Taking inverse sine transforms, we find u ( x,t ) = 2 π 0 T 0 w - T 0 w e - c 2 w 2 t sin wxdw = T 0 = sgn ( x )=1 2 π 0 sin wx w dw - 2 T 0 π 0 sin wx w e - c 2 w 2 t dw = T 0 - 2 T 0 π 0 sin wx w e - c 2 w 2 t dw 13. Proceed as in Exercise 11 using the Fourier sine transform instead of the cosine
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• Fall '11
• StuartChalk
• Inverse function, Discrete cosine transform, Integral transforms, Sine and cosine transforms, Discrete sine transform

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