Question

solve the differential equation using laplace transforms: y''' + 2y'' - y' - 2y = sin3t
Initial conditions: y(0) = 0, y'(0) = 0, y''(0) = 1

Answer 1

\[[s ^{3}Y - s ^{2}(y(0)) - s(y'(0)) - y''(0)] +2[s ^{2}Y - s(y(0)) - y'(0)] - [sY - y(0)] -2Y = 3/(s ^{2} +9)\]

\[s ^{3}Y + 2s ^{2}Y - 3sY = 3/(s ^{2}+9)\]

Answer 2

Sorry forgot about the -1 in the simplified equation from the initial condition y"(0)=1