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Differential Equations Lecture Work Solutions 76

# Differential Equations Lecture Work Solutions 76 - n=1 n x=...

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n =1 b n ( t ) sin n π L x = n =1 b n ( t ) n π L 2 sin n π L x + n =1 q n ( t ) sin We have a Fourier Sine series on left and Fourier Sine series on right, so the coeﬃcients must be the same; i.e., (a) b n ( t ) = n π L 2 b n ( t ) + q n ( t ) A first order ODE for b n ( t ). III. Solve b n ( t ) = n π L 2 b n ( t ) + q n ( t ) Solution Form: b n ( t ) = A n b H n ( t ) + b P n ( t ) Homogeneous Solution: b H n ( t ) = e ( n π L ) 2 t Particular Solution: b P n ( t ) = e ( n π L ) 2 t t 0 e ( n π L ) 2 τ q n ( τ ) b n ( t ) = A n e ( n π L ) 2 t + e ( n π L ) 2 t t 0 e ( n π L ) 2 τ q n ( τ ) (Step IV is an extra step, not required in homework problem.) IV . Find A n from initial condition. u ( x, 0) = f ( x ) = n =1 b n (0) sin n π L
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