Right, I understand the formula, but I thought that at the end, when I have completed the inverse Laplace and put the UnitStep function back into the formula for y(s), wherever there's a "t," I should put a "t-pi"? Which appears contradictory to placing "t+pi" here.
I too plugged it into Wolfram Alpha and got that result. I guess I can agree to that, though I still do not see the direct process used to derive the second and third terms.
If somebody can assist, that would be great.
Secondly, this is only half the problem - once we have this, we have to...
@ted s:
For the Right Hand Side, I got
L[coswt]+L[cos(ωt+π)], or s/(s^2+w^2) + L[cos(ωt+π)],
where cos(ωt+π), using the identity for cos(x+y), =cos(−ωt)
Therefore, the LHS is:
s/(s^2+w^2)+e^(−πs)⋅s/(s^2+ω^2)
Is that not correct?
I isolated F(s):
F(s) = [[s/(s²+w²)] + e^(−πs) * [s/ (s²+ω²)]] + 1 / ((s+2)²)
But I have no idea how to manipulate it to look like the ones in the table...
y′′+4y′+4y=f(t)
where f(t)=cos(ωt) if 0<t<π and f(t)=0 if t>π?
The initial conditions are y(0) = 0 , y'(0) = 1
I know that f(t)=cos(ωt)−uπ(t)cos(ωt), the heaviside equation.
AND ω is allowed to vary, supposed to find the general solution, i.e. f(t) in terms of ω
I think that after...
Use Laplace transfer to find the solution of the following initial value problem:
y''+4y'+4y=f(t)
where f(t) = cos(ωt) if 0<t<π and f(t)=0 if t>π ?
Also, y(0) = 0, y'(0) = 1
Currently, I have gotten to here, but not sure how to perform inverse Laplace:
(s+2)² * F(s) − 1 = [s/(s²+w²)]...