# Using Laplace Transforms to Solve PDE

ColdFusion85
Use the Laplace transform to solve

$$\frac{\partial^2 y }{ \partial t^2 } = c^2 \frac{ \partial^2 y }{ \partial x^2 }$$ for x>0, t>0

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

y(x,0) = 0, $$\frac{\partial y(x,0) }{ \partial t }$$ = 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

$$\frac{\partial^2 Y }{ \partial x^2 } = \frac{s^2}{c^2} Y - \frac{A}{c^2}$$

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

$$Y(x,s) = A(s)e^{\frac{s}{a} x} + B(s)e^{-\frac{s}{a} x}$$

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

$$Y(x,s) = B(s)e^{-\frac{s}{a} x}$$

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:

$$y(x,t) = At + (1-A)(t-\frac{x}{c})H(t-\frac{x}{c})$$

where H is the Heaviside function

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## Answers and Replies

Science Advisor
Homework Helper
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.

ColdFusion85
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.

ColdFusion85
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