Using Laplace Transform to Solve Non-Zero Initial Condition PDEs

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SUMMARY

The discussion focuses on solving a system of ordinary differential equations (ODEs) using the Laplace Transform method, specifically addressing non-zero initial conditions. The equations presented are x' - y = 1 and 2x' + x + y' = (t² - 2t + 1)e^-(t-1), with initial conditions x(1) = 0 and y(3) = 0. Key corrections include using L(x') = sX(s) - x(0) instead of X(s) - x(0) and properly transforming the non-homogeneous term (t² - 2t + 1)e^-(t-1). The solution involves leaving initial conditions as variables until the equations are solved for X(s) and Y(s).

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  • Understanding of Laplace Transform techniques
  • Familiarity with solving ordinary differential equations (ODEs)
  • Knowledge of initial value problems and their significance
  • Ability to perform inverse Laplace Transforms
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  • Study the properties of the Laplace Transform for initial value problems
  • Learn how to apply the inverse Laplace Transform to solve ODEs
  • Explore methods for handling non-homogeneous terms in differential equations
  • Review examples of systems of ODEs and their solutions using Laplace Transforms
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Students and educators in mathematics, particularly those studying differential equations, as well as engineers and scientists who apply Laplace Transforms in their work.

mike1111
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Homework Statement


Help, I don't know how to do the following question:

Using Laplace to solve
x' -y =1
2x' +x +y' = (t2-2t+1)e-(t-1)

Homework Equations


x(1)=0
y(3)=0


The Attempt at a Solution


The problem I'm having is the initial conditions aren't at zero, and I'm not sure how to approach the question

so far I have:
X(s) -x(0) -Y(s) = 1/s
3X(s) -2x(0) +Y(s)-y(0)= F{(t2-2t+1)e-(t-1)}
 
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This isn't a pde. It is a system of ordinary DE's. And you have mistakes in your transforms.

L(x') is not X(s) - x(0) it is sX(s) - x(0), and ditto for y'. And, of course, you need to transform (t2-2t+1)e-(t-1).

Just call x(0) = a and y(0) = b and leave them in there. Once you get the equations right and solve for X(s) and Y(s), you can take the inverse transforms. Your answers will have a and b in them. Finally, plug in your given initial conditions and choose a and b to make them work.
 
Thanks a lot LCKurtz, I didn't see that mistake. and the solution makes more sense now. I was getting weird answer from other questions too and could work out why.
 

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