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Using Laplace Transforms to Solve PDE

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Use the Laplace transform to solve

[tex]\frac{\partial^2 y }{ \partial t^2 } = c^2 \frac{ \partial^2 y }{ \partial x^2 }[/tex] for x>0, t>0

y(0,t) = t, for t>0

y(x,0) = 0, [tex]\frac{\partial y(x,0) }{ \partial t }[/tex] = A, for x>0


So I used the Laplace transform of a derivative, along with the initial conditions to get the PDE in the form

[tex]\frac{\partial^2 Y }{ \partial x^2 } = \frac{s^2}{c^2} Y - \frac{A}{c^2}[/tex]

However, I am not sure how to solve this. I know if the A/c^2 wasn't there it would just be

[tex] Y(x,s) = A(s)e^{\frac{s}{a} x} + B(s)e^{-\frac{s}{a} x}[/tex]

and since Y must be finite for all s because the Laplace transform of a finite function is finite, A(s)=0, leaving just

[tex] Y(x,s) = B(s)e^{-\frac{s}{a} x}[/tex]

and my notes from class tell me how to proceed from here. My problem is I don't know how to generate a form of the solution to my problem that I arrived at above. Can anyone help me proceed from here?


If it is any help, the book tells me the final answer should be:

[tex]y(x,t) = At + (1-A)(t-\frac{x}{c})H(t-\frac{x}{c})[/tex]

where H is the Heaviside function
 
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Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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You are using the Laplace transform on partial differential equations and don't know how to handle a non-homogeneous linear equation with constant coefficients? It must have be been a long time since you took an introductory course in ordinary differential equations!

Use "undetermined coefficients" with y= C, constant. Put that into the equation to see what C must be in order to satisfy the equation.
 
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This doesn't work. If I guess y=C, then we find that C=A/s^2. Y(0,s) = t, and taking the Laplace transform if this gives us 1/s^2, meaning A=1. So Y(x,s) = 1/s^2 and therefore y(x,t) = t. This isn't even close to the answer in the book.
 
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Nevermind, I solved the problem. It involves an exponential term plus an A/s term prior to taking the inverse transform to arrive at the solution. Guessing y=C wasn't the correct method. Thanks anyway though
 

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