Recent content by jakejakejake

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.

OK I almost got it. One last thing: Why is it cos(ω(t+π)) and not cos(ω(t-π))?

Right, but that just means the second term is e^(−πs)*L[cos(ωt+π)], where cos(ωt+π), using the identity for cos(x+y), = cos(−ωt), no?

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...

Sorry - Edited!

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)]...