IVP Laplace Transform Problem - Tricky Inverse Laplace Transform

[jimagnus]

IVP Laplace Transform Problem -- Tricky Inverse Laplace Transform

Homework Statement

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

Homework Equations

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?

 

[MisterX]

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

 

[vela]

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

Why not?

 

[jimagnus]

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.

 

[vela]

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

 

[jimagnus]

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!