Using Laplace transforms to solve differential equations - with a twist

Click For Summary

Homework Help Overview

The problem involves solving the differential equation x'' + x = 4δ(t-2π) using Laplace transforms, with initial conditions x(0) = 1 and x'(0) = 0. The challenge is to express the final solution without using the unit step function.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the transformation of the equation using Laplace transforms and express confusion about how to present the solution without the unit step function. There is consideration of the periodic nature of the sine function and whether a simpler form can be used.

Discussion Status

Some participants have provided insights into expressing the solution as a piecewise function, while others question if there are alternative methods to approach the problem. There is an acknowledgment of the need to incorporate initial conditions, but no consensus on the final form of the solution has been reached.

Contextual Notes

Participants note the specific requirement to avoid the unit step function in the final answer, which adds complexity to the problem. There is also a mention of the lecturer's expectations, suggesting that the problem may have nuances that are not immediately apparent.

Laura W
Messages
2
Reaction score
0
I've been given this:
x''+ x = 4δ(t-2π)

The question asks:
With initial conditions of x(0) = 1 and x'(0) = 0, find x(t) using Laplace transforms.

I can easily get it to this:

4(sin(t-2π)u(t-2π))

But the question says "express your final solution without use of the unit step function". This is where I get confused as I'm not quite sure as to how to do that. Will it just be 4sin(t)? Considering the sin function repeats at 2π.
 
Physics news on Phys.org
Laura W said:
I've been given this:
x''+ x = 4δ(t-2π)

The question asks:
With initial conditions of x(0) = 1 and x'(0) = 0, find x(t) using Laplace transforms.

I can easily get it to this:

4(sin(t-2π)u(t-2π))

But the question says "express your final solution without use of the unit step function". This is where I get confused as I'm not quite sure as to how to do that. Will it just be 4sin(t)? Considering the sin function repeats at 2π.

In general if you have ##x=f(t)u(t-a)## you would write a two piece function since ##u(t-a) = 0## if ##t<a## and ##u(t-a) = 1## if ##t>a##. So you would say$$
x =\left\{\begin{array}{l}
0,~t<a\\
f(t),~t>a
\end{array}\right.$$Does that help?
 
LCKurtz said:
In general if you have ##x=f(t)u(t-a)## you would write a two piece function since ##u(t-a) = 0## if ##t<a## and ##u(t-a) = 1## if ##t>a##. So you would say$$
x =\left\{\begin{array}{l}
0,~t<a\\
f(t),~t>a
\end{array}\right.$$Does that help?

That's what I was thinking but I was wondering if there was another way of transforming it at all? Seems to simple of an answer for this specific lecturer who wrote the question.
 
Laura W said:
I've been given this:
x''+ x = 4δ(t-2π)

The question asks:
With initial conditions of x(0) = 1 and x'(0) = 0, find x(t) using Laplace transforms.

I can easily get it to this:

4(sin(t-2π)u(t-2π))

But the question says "express your final solution without use of the unit step function". This is where I get confused as I'm not quite sure as to how to do that. Will it just be 4sin(t)? Considering the sin function repeats at 2π.
Your answer isn't correct. Did you forget to incorporate the initial conditions?
 

Similar threads

Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K