Laplace Transform/Diff eq question

[smk037]

Homework Statement

Find the solution to the differential equation using laplace transform techniques

y''(t) + 2y'(t) + 2y(t) = u'(t)
u(t) = 1(t) (step function)
y(0) = 1, y'(0) = 0

Homework Equations

L{y''(t)} = (s^2)Y(s) + (s)y(0) - y'(0)
L{y'(t)} = (s)Y(s) - y(0)
u'(t) = d(t) (dirac/delta function)
L{d(t)} = 1

The Attempt at a Solution

L{y''(t) + 2y'(t) + 2y(t) = u'(t)} =
(s^2)Y(s) + (s)y(0) - y'(0) - 2s*Y(s) - 2y(0) + 2Y(s) = 1
(s^2)Y(s) - s + 2s*Y(s) - 2 + 2Y(s) = 1
(s^2)Y(s) + 2s*Y(s) + 2Y(s) = 3+s
Y(s) (s^2 + 2s + 2) = 3+s
Y(s) = (3+s)/(s^2 + 2s + 2)

Now, the problem I am facing is that any calculator that I use tells me the solution to this problem is (e^(-t))*(cos(t) + sin(t))
example
http://www.wolframalpha.com/input/?i=y''(t)+2y'(t)+2y(t)=+dirac(t),+y(0)+=+1,+y'(0)+=+0

I realized this when doing my analysis of the system in MATLAB.

But when I take the inverse laplace transform of the equation I ended up with, I get
(e^(-t))*(cos(t) + 2sin(t))

which is inconsistent with the initial conditions, since it gives a 2 at t = 0.
Any ideas about what I could be doing wrong here?

 

[NJunJie]

e^(-0)=1
cos(0) =1
2sin(0) =0

so y(0)=1.

Your working seems correct :)