Laplace Transform Solution for x"+2x'+x=sin(t) x=x'=0

[jumpman23bd]

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.

 

[Unco]

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 $$\frac{s}{(s+1)^2+4}$$ using the formulae

$$\frac{s-a}{(s-a)^2+b^2}\mapsto e^{at}\cos{bt}$$

and

$$\frac{b}{(s-a)^2+b^2}\mapsto e^{at}\sin{bt}\, ,$$

begin by rewriting it as

$$\frac{s}{(s+1)^2+4} = \frac{s+1}{(s+1)^2+4} - \frac{1}{(s+1)^2+4}$$.

Can you proceed from there?

 

[jumpman23bd]

yes I do see it now...thank you