Solving Laplace Transforms: y""-4y"'+6y" -4y'+y=0

[hbomb]

Can someone show me how to do these laplace transforms of these differentials?

1) y""-4y"'+6y" -4y'+y=0
y(0)=0, y'(0)=1, y"(0)=0, y"'(0)=1

2) y"-2y'+4y=0
y(0)=2, y'(0)=0

3) y"'+2y'+y=4e^-t
y(0)=2, y'(0)=-1

4) y"-2y'+2y=cos(t)
y(0)=1, y'(0)=0
the Laplace transfrom that i got for this was
s/(s^2+a^2) * 1/(s^2-2s+2) + (s-2)/(s^2-2s+2)=y
I'm trying to find the inverse transforms of these but i have no idea how to do this because i can't factor the numerator by completing the square.

 

[HallsofIvy]

Are you trying to find the Laplace transform or the inverse transform?

Surely, if you are doing problems like that, you must know that:
L(y')= sL(y)- y(0),
L(y")= s²L(y)- y(0)- y'(0), and
L(y"')= s³L(y)- y(0)- y'(0)- y"(0).

s²- 2x+ 2= s²-2x+ 1+ 1= (s-1)²+ 1. You can't factor that, of course (with real numbers), but you should know inverse transforms involving $\frac{1}{s^2+ 1}$.

 

[hbomb]

My mistake, I'm looking for the inverse Laplace transform.