Ordinary Differential Equations

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Homework Help Overview

The discussion revolves around finding the general solution to an initial value problem (IVP) involving an ordinary differential equation (ODE) with a delta function as a forcing term. The problem specifically addresses the behavior of the solution at a point of discontinuity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to derive the jump condition for the solution and discuss the implications of integrating the delta function. Questions arise regarding the interpretation of the results from integration and the nature of the discontinuity in the solution.

Discussion Status

Some participants have suggested a method of integrating the differential equation around the point of discontinuity to determine the size of the jump in the solution. There is ongoing exploration of the implications of integrating the delta function and its effect on the solution.

Contextual Notes

The original poster notes a lack of clarity regarding the jump condition as it was not covered in lectures or the textbook, indicating a gap in the provided resources for understanding this aspect of the problem.

sassie
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Homework Statement



Give the general solution to the IVP

L[y]=y'+(sint)y=[tex]\delta[/tex](t-[tex]\tau[/tex])
y(0)=0

For all t>0 by placing a jump condition on y(t) and solving the differential equation for t<[tex]\tau[/tex] and t>[tex]\tau[/tex]

Homework Equations


The Attempt at a Solution



I'm plenty sure I can get the general solution to the problem, but I do not at all know how to get the "jump condition" as it wasn't explained in lectures or in the textbook. Your help is very much appreciated.
 
Last edited:
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Integrate the differential equation from [itex]\tau-\varepsilon[/itex] to [itex]\tau+\varepsilon[/itex] and take the limit as [itex]\varepsilon\rightarrow 0[/itex]. That will give you a result that tells you how big the discontinuity in y(t) is at [itex]t=\tau[/itex].
 
vela said:
Integrate the differential equation from [itex]\tau-\varepsilon[/itex] to [itex]\tau+\varepsilon[/itex] and take the limit as [itex]\varepsilon\rightarrow 0[/itex]. That will give you a result that tells you how big the discontinuity in y(t) is at [itex]t=\tau[/itex].

So what would we get when we integrate [tex]\delta[/tex](t-[tex]\tau[/tex])? would it be 1?
 
Yes, because the interval of integration includes the point [itex]t=\tau[/itex].
 

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