To solve the given differential equation using Laplace transform, we first take the Laplace transform of both sides with respect to t:
L[G(t)] + L[Y(t) cosh(t-T)] = L[t] + L[E(t)]
where L denotes the Laplace transform operator.
Using the properties of Laplace transform, we have:
L[G(t)] = G(s) (where s is the Laplace variable)
L[Y(t) cosh(t-T)] = Y(s) * L[cosh(t-T)] = Y(s) * [s/(s^2-1)]
L[t] = 1/s^2
L[E(t)] = E(s)
Substituting these values into the equation and solving for Y(s), we get:
Y(s) * [s/(s^2-1)] + G(s) = 1/s^2 + E(s)
Y(s) = [1/s^2 + E(s) - G(s)] * [(s^2-1)/s]
To obtain the solution in the time domain, we use partial fraction decomposition and inverse Laplace transform:
Y(s) = [A/s + B/s^2 + C/(s+1) + D/(s-1)]
where A = lim(s->0) sY(s), B = lim(s->0) ds/dt[Y(s)], C = Y(-1), D = Y(1).
After finding the values of A, B, C, and D, we can write the solution in the time domain as:
Y(t) = A + Bt + Ce^(-t) + De^(t)
where e^(-t) and e^(t) are the inverse Laplace transforms of 1/(s+1) and 1/(s-1), respectively.
Note that the Laplace transform solution assumes that the initial conditions are zero. If there are non-zero initial conditions, they need to be incorporated into the solution.