Laplace transform

  • #1

Homework Statement


Problem: laplace transform of x"+2x'+x=sin(t) x=x'=0


Homework Equations





The Attempt at a Solution


Attempt at problem: I was able to get Y(s)=1/(s^2+1)(s^2+2s+5) and then i expanded it to get [s/10(s^2+2s+5)]-[s/10(s^2+1)]+[1/5(s^2+1)]. The last two terms are easy to transform the first term i completed the square and got 1/10*s/(s+1)^2+4

Question: from maple i know it will be a exp times a sin and cos function of some sort. I don't know how they got that even though i see it in the denominator.
 

Answers and Replies

  • #2
156
0
Hi Jumpman,

Firstly, if you did mean to write the DE as x"+2x'+x=sin(t) (with no 5 in front of the x) then you should restart from there. Otherwise, I'll assume you meant x"+2x'+5x=sin(t) with initial conditions x(0)=x'(0)=0.

To find the inverse laplace of [tex]\frac{s}{(s+1)^2+4}[/tex] using the formulae

[tex]\frac{s-a}{(s-a)^2+b^2}\mapsto e^{at}\cos{bt}[/tex]

and

[tex]\frac{b}{(s-a)^2+b^2}\mapsto e^{at}\sin{bt}\, ,[/tex]

begin by rewriting it as

[tex]\frac{s}{(s+1)^2+4} = \frac{s+1}{(s+1)^2+4} - \frac{1}{(s+1)^2+4}[/tex].

Can you proceed from there?
 
  • #3
yes I do see it now...thank you
 

Related Threads on Laplace transform

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
504
  • Last Post
Replies
1
Views
741
  • Last Post
Replies
3
Views
972
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
702
Top