Laplace Transform with IVP and completing the square

[sami23]

Homework Statement

Use the Laplace Transform to solve: y"+2y'+2y=t y(0)=y'(0)=1

Homework Equations

L{y(t)} = Y(s)
L{y'(t)} = sY(s)-y(s)
L{y"(t)} = s²Y(s)-sy(0)-y'(0)
using the laplace transform table: tⁿ = n!/(sⁿ⁺¹) where n=1

The Attempt at a Solution

Take laplace on both sides:
L{y"(t)} + 2L{y'(t)} + 2L{y(t)} = L{t}
s²Y(s) - sy(0) - y'(0) + 2[sY(s)-y(0)] + 2Y(s) = 1/s²

after plugging in the initial conditions I get:
s²Y(s) - s - 3 + 2sY(s) + 2Y(s) = 1/s²

isolate Y(s):
Y(s)[s²+2s+2] = 1/s² + s + 3
Y(s) = (1/s² + s + 3) / (s²+2s+2)
I completed the square in the denominator to get: (m+1)²+1

Y(s) = s / [(s²+1)(s+3)²-8]

Take the Laplace inverse
L^-1{ s / [(s²+1)(s+3)²-8]}

I add and subtract 3 in the numerator to get:
L^-1{ (s+3)-3 / [(s²+1)(s+3)²-8]}

Use linearity property of inverse transform to get from L^-1{Y(s)} to y(t):
L^-1{ (s+3) / [(s²+1)(s+3)²-8]} - 3L^-1{ 1 / [(s²+1)(s+3)²-8}

How do I apply partial fraction decomposition to get y(t)?

 

[jweygna1]

Right after you completed the square in the denominator, I would have split it up there to:
Y(s)=1/(s^2(s^2+2s+2)) + (s+3)/(s^2+2s+2) = u+v
There we can see that
v=(s+1)/(s^2+2s+2) +2/(s^2+2s+2) so
L-1{v}=e^(-t)(cos(t)+2sin(t))
Now partial fractions with just u= 1/(s^2(s^2+2s+2)) is easier than what you tried to do.
Set it up as:
u=A/s+B/s^2+(Cs+D)/1/(s^2+2s+2). Now find A,B,C,and D and you'll be home free.

 

[jweygna1]

Right after you completed the square in the denominator, I would have split it up there to:
Y(s)=1/(s^2(s^2+2s+2)) + (s+3)/(s^2+2s+2) = u+v
There we can see that
v=(s+1)/(s^2+2s+2) +2/(s^2+2s+2) so
L-1{v}=e^(-t)(cos(t)+2sin(t))
Now partial fractions with just u= 1/(s^2(s^2+2s+2)) is easier than what you tried to do.
Set it up as:
u=A/s+B/s^2+(Cs+D)/(s^2+2s+2). Now find A,B,C,and D and you'll be home free.

 

[sami23]

Thank you so much! Made my life easier.