# The Dirac Delta Function

## Homework Statement

Differential equation: ##Ay''+By'+Cy=f(t)## with ##y_{0}=y'_{0}=0##

Write the solution as a convolution (##a \neq b##). Let ##f(t)= n## for ##t_{0} < t < t_{0}+\frac{1}{n}##. Find y and then let ##n \rightarrow \infty##.

Then solve the differential equation with ##f(t)=\delta(t-t_{0})##.

## Homework Equations

Convolution (Boas)
Laplace Transforms (Boas)

## The Attempt at a Solution

So when I go through the first part with ##f(t)= n## for ##t_{0} < t < t_{0}+\frac{1}{n}## and do convolution I get ##y=\frac{1}{A(b-a)}\int_0^t(e^{-a(t-\tau)}-e^{-b(t-\tau)})f(\tau)d\tau## which has different cases depending on t:

0 if ##t<t_{0}##

##\frac{n}{A(b-a)}(\frac{1}{a}(1-e^{a(t_{0}-t)})-\frac{1}{b}(1-e^{b(t_{0}-t)}))## if ##t_{0}<t<t_{0}+\frac{1}{n}##

##\frac{n}{A(b-a)}(\frac{1}{a}(e^{-a(t-t_{0}-\frac{1}{n})}-e^{a(t_{0}-t)})-\frac{1}{b}(e^{-b(t-t_{0}-\frac{1}{n})}-e^{b(t_{0}-t)}))## if ##t>t_{0}+\frac{1}{n}##

I don't understand what happens as ##n\rightarrow \infty##. I know it should become

##\frac{1}{A(b-a)}(e^{a(t_{0}-t)}-e^{b(t_{0}-t)})## for ##t>t_{0}## because that is what I get when I use ##f(t)=\delta(t-t_{0})## from the beginning. But what happens to all the extra terms? And the n out front that goes to infinity?

## Answers and Replies

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## Homework Statement

Differential equation: ##Ay''+By'+Cy=f(t)## with ##y_{0}=y'_{0}=0##

Write the solution as a convolution (##a \neq b##). Let ##f(t)= n## for ##t_{0} < t < t_{0}+\frac{1}{n}##. Find y and then let ##n \rightarrow \infty##.

Then solve the differential equation with ##f(t)=\delta(t-t_{0})##.

## Homework Equations

Convolution (Boas)
Laplace Transforms (Boas)

## The Attempt at a Solution

So when I go through the first part with ##f(t)= n## for ##t_{0} < t < t_{0}+\frac{1}{n}## and do convolution I get ##y=\frac{1}{A(b-a)}\int_0^t(e^{-a(t-\tau)}-e^{-b(t-\tau)})f(\tau)d\tau## which has different cases depending on t:

0 if ##t<t_{0}##

##\frac{n}{A(b-a)}(\frac{1}{a}(1-e^{a(t_{0}-t)})-\frac{1}{b}(1-e^{b(t_{0}-t)}))## if ##t_{0}<t<t_{0}+\frac{1}{n}##

##\frac{n}{A(b-a)}(\frac{1}{a}(e^{-a(t-t_{0}-\frac{1}{n})}-e^{a(t_{0}-t)})-\frac{1}{b}(e^{-b(t-t_{0}-\frac{1}{n})}-e^{b(t_{0}-t)}))## if ##t>t_{0}+\frac{1}{n}##

I don't understand what happens as ##n\rightarrow \infty##. I know it should become

##\frac{1}{A(b-a)}(e^{a(t_{0}-t)}-e^{b(t_{0}-t)})## for ##t>t_{0}## because that is what I get when I use ##f(t)=\delta(t-t_{0})## from the beginning. But what happens to all the extra terms? And the n out front that goes to infinity?
Is ##f(t)## supposed to be zero for ##t < t_0## and ##t > t_0+\frac{1}{n}##?

Is f(t)f(t)f(t) supposed to be zero for t<t0t<t0t < t_0 and t>t0+1nt>t0+1nt > t_0+\frac{1}{n}?
Yes! Sorry I forgot to specify that.