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

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SUMMARY

The discussion centers on solving the differential equation x'' + 2x' + x = sin(t) with initial conditions x(0) = 0 and x'(0) = 0 using the Laplace Transform. The participant successfully derived Y(s) = 1/[(s^2 + 1)(s^2 + 2s + 5)] and expanded it into simpler components for inverse transformation. The key formulas for inverse Laplace transformation were identified, specifically the forms for e^(at)cos(bt) and e^(at)sin(bt), which are crucial for deriving the solution.

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jumpman23bd
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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.
 
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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?
 
yes I do see it now...thank you
 

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