Solving Impulse-Diffy Equation: y''+2y'+3y=sin(t)+δ(t-3π)

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SUMMARY

The discussion focuses on solving the Impulse-Diffy Equation represented as y'' + 2y' + 3y = sin(t) + δ(t - 3π). The left side of the equation is expressed as Y(s)(s² + 1) - 1, indicating a transformation into the Laplace domain. The challenge arises in handling the delta function, which is addressed using the property L[δ(t - t₀)] = e^{-t₀ s}. This approach is essential for integrating the delta function effectively in the context of the equation.

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  • Understanding of Laplace transforms
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  • Basic integration techniques in calculus
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Homework Statement


y''+2y'+3y=sin(t)+\delta (t-3 \pi )


Homework Equations





The Attempt at a Solution


Left side is just Y(s)*(s^2+1)-1
But I don't know how to deal with the delta function, I made it into just an intergral but I don't know how to intergrate it.
 
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L[\delta(t-t_0)]=e^{-t_0 s}


If I remember correctly that is.
 
thanks=)
 

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