Question

help with this linear algebra question

dy =1 +1 dt X-- Solve the system: 5.) dx dy 2y=e + dt d

Answer 1

x - d^2 y/dt^2 = t + 1 (1) dx/dt + dy/dt - 2y = e^t (2) Rightarrow dx/dt - d^3 y/dt^3 = 1 Rightarrow dx/dt = 1 + d^3y/dt^3 Rightarrow d^3y/dt^3 + dy/dt - 2y = e^t - 1 = e^t - e^0 A E: m^3 + m - 2 = 0 Rightarrow (m - 1) (m^2 + m + 2) = 0 Rightarrow m = 1 (or) m = -1/2 plusminus 7/2 i Therefore y_c = c_1 e^t + e^- 1/2t[c_1 cos (squareroot 7/2 t) c_2 sin (squareroot 7/2 t)] y_D = e^t/D^3 + D - 2 - e^0/D^3 + D - 2 = t^e^t/3D^2 + 1 - e^0/D^3 + D + 2 (because D^3 + D - 2 = 0 if D = 1) t e^t/3 (1)^2 + 1 - e^0/0^3 + 0 - 2 = t/4 e^t + 1/2 Therefore y = y_c + y_p = c_1 e^t + e^- 1/2t [c_1 cos (squareroot 7/2 t) + c_2 sin squareroot 7/2 t)] + t/4 e^t + 1/2 x = t + 1 + d^2y/dt^2 = t + 1 + c_1 e^t + 7/4 e^- 1/2t [- c_1 cos (squareroot 7/2 t) + c_2 sin (squareroot 7/2 t)] + 1/4 (t - 2) e^t