homework question

[frontwards]

i'm having a tough time understanding this problem and i thought maybe someone could help me out. this is what i am having trouble with....

why does L^-1{(1/5)(e^-s)(1/(s+1)-(1/5)(e^-s)(s/((s^2)+4))+(1/5)(e^-s)(1/((s^2)+4)) = (1/5)u(x-1)(e^-(x-1))-(1/5)u(x-1)cos(2x-2)+(1/10)u(x-1)sin(2x-2)

i'm trying to understand how all of that is gotten using a table of general laplace transforms, but i'm just not seeing it. my prof did this problem in class today and he said we could do it using a table but i am at a loss understanding which tables he used and why. hopefully someone has time for this. thanks.

 

[Krakerjak]

Well, I will factor out the (1/5)e^-s from all groups

(0.2e^-s) * ( (1/(s+1)) - (s/(s^2+2^2)) + (1/(s^2+2^2)) )

You can just look up laplace transform tables on google..... http://www.vibrationdata.com/Laplace.htm

According to the table, the inverse laplace transform of 0.2 e^-s -----------> 0.2 u(x-1)

Now, I'm pretty rusty with this, I can't quite figure out how the other functions wind up with (t + ...) someone else may know, but this should give you an idea.

1/(s+1) ---------> e^-x

s/(s^2+2^2) cos(2x)

1/(s^2+2^2) ---------> 0.5 * (2/(s^2+2^2)) -----------> 0.5 * sin(2x)

put it together and you have

0.2*u(x-1)(e^-(x-1)) - 0.2*u(x-1)cos(2x-2) + 0.1*u(x-1)sin(2x-2)