How can I use the Dirac delta function in a Laplace transform?

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Homework Help Overview

The problem involves a second-order linear differential equation with a Dirac delta function on the right-hand side, specifically in the context of Laplace transforms. The equation is given as 2y'' + y' + 4y = ∂(t-π/6)sin(t), with initial conditions y(0)=0 and y'(0)=1/2.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Dirac delta function in the context of the Laplace transform, questioning how to handle the right-hand side of the equation. There is mention of evaluating integrals involving the delta function and its properties.

Discussion Status

The discussion is active, with participants providing insights on how to approach the problem. Some suggest evaluating the integral directly using the definition of the Laplace transform, while others highlight the properties of the Dirac delta function as a means to simplify the process.

Contextual Notes

There is a noted uncertainty regarding the application of the Dirac delta function in conjunction with other functions, as well as the interpretation of the right-hand side of the equation. Participants are navigating these aspects without reaching a consensus.

Jamin2112
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Homework Statement



2y'' + y' + 4y = ∂(t-π/6)sin(t); y(0)=0, y'(0)=1/2.

Homework Equations



Dunno

The Attempt at a Solution



The left side of the equation is what's tripping me up. There's nothing on my Elementary Laplace Transforms table that has the Dirac delta function multiplied by another function, or sin(at) multiplied by another function f(t). So what should I do?
 
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You mean the right side? The delta functions makes it very easy actually. Consider the value of the integral when [itex]t \neq \pi/6[/itex]. Now consider [itex]t = \pi/6[/itex].
 
I think you meant the right-hand side. Anyway, just plug the RHS into the definition of the Laplace transform and evaluate the integral. The delta function makes it easy to do.
 
One of the properties of the Dirac delta is that for any sufficiently smooth function [itex]f[/itex], [itex]\int f(x)\delta(x-a)dx=f(a)[/itex] if the integration interval includes the point [itex]x=a[/itex], and the integral is zero otherwise. Use that.
 

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