Solving Differential Equation w/ Laplace Transform

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SUMMARY

The discussion focuses on solving a differential equation using the Laplace transform, specifically addressing the inverse transform of the expression (exp(-s))/(s^2). The user initially struggled to find this expression in Laplace Transform Tables but received guidance on using the Residue theorem for evaluation. The final result is f(t) = 0 for t < 1 and f(t) = t/2 for t > 1, indicating the behavior of the function based on the semi-circular contour chosen in the complex plane.

PREREQUISITES
  • Understanding of Laplace transforms and their properties
  • Familiarity with differential equations
  • Knowledge of complex analysis, specifically the Residue theorem
  • Experience with contour integration techniques
NEXT STEPS
  • Study the application of the Residue theorem in complex analysis
  • Learn about different types of poles and their contributions in contour integrals
  • Explore advanced Laplace transform techniques and their applications
  • Review differential equations and their solutions using Laplace transforms
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Students and professionals in mathematics, engineering, and physics who are working on differential equations and require a solid understanding of Laplace transforms and complex analysis techniques.

Neoon
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Hi all,

Just a small question:
I came across a problem to solve a differential equation using Laplace transform.

I solved the major part but only still a part where I had to inverse transform the following experssion:

(exp(-s))/(s^2)

I looked in Laplace Transform Tables but did not find it.

Could you please help?
 
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think i got it..

you're looking for the integral..

[tex](2\pi i)^{-1} \oint ds exp(-st-s)/s^{2}[/tex]

now using 'Residue theorem' you get...

[tex]f(t)=0[/tex] iff t<1 and f(t)=t/2 [/tex] iff t>1.


depending on if you take the semi-circle (-T,0) or (0,T) with T--->oo , there is a double pole at s=0
 

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