Using Laplace Transform to solve a differential equation

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SUMMARY

The discussion focuses on solving the differential equation y" + y = 4δ(t-2π) with initial conditions y(0)=1 and y'(0)=0 using the Laplace Transform method. The user derived the solution as cos(t) + 4U(t-2π)sin(t-2π), while WolframAlpha provided a different result of 4sin(t)U(t-2π) + cos(t). The discrepancy arises from the periodic nature of the sine function, specifically that sin(t-2π) simplifies to sin(t).

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with Laplace Transform techniques
  • Knowledge of the Dirac delta function
  • Basic trigonometric identities and properties
NEXT STEPS
  • Review the properties of the Laplace Transform, specifically L[δ(t-c)] = e^{-cs}
  • Study the periodic properties of trigonometric functions, particularly sine and cosine
  • Practice solving differential equations using the Laplace Transform method
  • Explore the use of WolframAlpha for verifying mathematical solutions
USEFUL FOR

Students studying differential equations, mathematicians interested in Laplace Transforms, and anyone seeking to clarify trigonometric identities in the context of differential equations.

november1992
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Homework Statement



y" + y = 4δ(t-2π); y(0)=1, y'(0)=0

Homework Equations



L[f(t-a) U(t-a)] = [itex]e^{-as}[/itex] L[f(t)]

L[δ(t-c)] = [itex]e^{-cs}[/itex]

The Attempt at a Solution



My answer is: cos(t) + 4U(t-2π)sin(t-2π).

When I used Wolframalpha it gave me 4sin(t)U(t-2π) + cos(t)
 
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Hi november1992!

So what is your question?

Did you know that sin(t-2π)=sin(t) since the sine has a period of 2π?
 
I meant to ask why is my answer different. I don't remember any of the trig identities so I'll have to review them.

Thanks for answering my question!
 

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