IVP Laplace Transform Problem - Tricky Inverse Laplace Transform

1. Jan 29, 2012

jimagnus

IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

1. The problem statement, all variables and given/known data

Solve x"+x'+x=1, given x(0)=x'(0)=0

2. Relevant equations

3. The attempt at a solution

Plugged in transforms: s2*Y(s)-s*y(0)-y'(0)+s*Y(s)-y(0)+Y(s)=1/s
Plugged in initial value points, simplified to Y(s)=1/((s2+s+1)*s)
Partial Fractions led me to Y(s)=1/s+ (-s-1)/(s^2+s+1)

I get stuck at finding an inverse transform for the second term. You can't complete the square for the denominator, right?

2. Jan 29, 2012

MisterX

Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

You can still do a partial fraction expansion even if the roots of the denominator aren't real.

Last edited: Jan 29, 2012
3. Jan 29, 2012

vela

Staff Emeritus
Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

Why not?

4. Jan 30, 2012

jimagnus

Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

If I complete the square, I get the denominator is (s+1/2)2+3/4, and the numerator is -(s+1), and I don't know what inverse laplace transform this would be.

5. Jan 30, 2012

vela

Staff Emeritus
Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

Hint: s+1 = (s+1/2) + 1/2

6. Jan 30, 2012

jimagnus

Re: IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

oh right, I didn't think to break it up like that.

so you end up with 1/s-[(s+1/2)/((s+1/2)2+3/4)+1/2 / ((s+1/2)2+3/4), which becomes 1-e-t/2cos3t/4-√3/3 * e-t/2sin√3/2

thanks!