"

Section 4.5 Impulse functions

Objective:

1. Definition of the impulse function

2. Use Laplace transformation to solve differential equations with impulse function

We are introducing a new function, impulse functions. An electrical circuit or mechanical system subject to a sudden voltage or force g(t) of large magnitude that acts over a short time interval about t=c behave as impulse functions. Those are discontinue functions.

 

Definition: The unit impulse function δ is defined to have the properties δ(t)=0 for t0, and δ(t)dt=1. In general, for a unit impulse at an arbitrary point c, δ(tc)=0 for tc, and δ(tc)dt=1.

 

 

Theorem: Laplace Transform of δ(tc) is ecs. The proof for δ(t), i.e. c=0 is at the page 32, 33, 34, and 35 here. 

 

 

Example 1: Use Laplace Transform to solve the IVP. y+4y+8y=δ(t2) y(0)=1, y(0)=1. 

 

 

 

Exercise 1: Use Laplace Transform to solve the IVP. y+2y+5y=δ(t1), y(0)=1, y(0)=2. 

 

 

 

Example 2: Use Laplace Transform to solve the IVP. y+9y=δ(t4)δ(t8)y(0)=0, y(0)=1. 

 

 

 

Exercise 2: Use Laplace Transform to solve the IVP. y+4y=δ(t3)+δ(t2), y(0)=1, y(0)=0. 

 

 

 

Example 3: Use Laplace Transform to solve the IVP. yy2y=δ(t3)+u4(t)y(0)=0, y(0)=1. 

 

 

 

Exercise 3: Use Laplace Transform to solve the IVP. y+2y8y=δ(t5)u3(t), y(0)=1, y(0)=1. 

 

 

 

Example 4: Use Laplace Transform to solve the IVP. y+2y+3y=δ(t3π)+sin(t)y(0)=0, y(0)=0. 

 

 

 

Exercise 4: Use Laplace Transform to solve the IVP. y+4y+13y=δ(tπ)cos(t), y(0)=0, y(0)=0. 

 

 

 

Group Work

1. Use Laplace Transform to solve the IVP. y+2y+5y=u2(t)3δ(t4)y(0)=6, y(0)=0. 

 

2. Use Laplace Transform to solve the IVP. y+4y+4y=e2tδ(te)y(0)=0, y(0)=1. 

License

Differential Equations Copyright © by Kuei-Nuan Lin. All Rights Reserved.