Solve the differential equation y" - 18y' = 8(t + 3) + 28'(t) Where the initial conditions are y(0) = 0, y'(0) = 6 Note: You can use the result L{8'(t) } = s, if required in your calculations. Solution: 1 1 7 Y(s) = es + 18(s - 18) 18s. 1 - - 3(s18) 3s The inverse Laplace transform and hence y(t) becomes: 18(t+3) u(t+3) y(t) = u(t + 3) + 18 18 7e18t 3 - 1 u(t)