Ordinary Differential Equations

sassie
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Homework Statement



Give the general solution to the IVP

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

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

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.
 
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Integrate the differential equation from \tau-\varepsilon to \tau+\varepsilon and take the limit as \varepsilon\rightarrow 0. That will give you a result that tells you how big the discontinuity in y(t) is at t=\tau.
 
vela said:
Integrate the differential equation from \tau-\varepsilon to \tau+\varepsilon and take the limit as \varepsilon\rightarrow 0. That will give you a result that tells you how big the discontinuity in y(t) is at t=\tau.

So what would we get when we integrate \delta(t-\tau)? would it be 1?
 
Yes, because the interval of integration includes the point t=\tau.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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