laplace transforms

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 [itex]\frac{1}{s^2+ 1}[/itex].

hbomb

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