PDE 83215 Solutions to exercises 5 1. (a) The general solution of utt ...

4. (a) To make the boundary conditions homogeneous we writeThe function v must satisfyWe look for a solution in the formu(x, t) = xt + v(x, t)v tt − v xx = cos 2t cos 3xv x (0, t) = v x (π, t) = 0v(x, 0) = cos 2 xv t (x, 0) = 0v(x, t) = T 0 (t) + T 2 (t) cos 2x + T 3 (t) cos 3xThe boundary conditions are satisfied. The differential equation givesand the initial conditions giveSolving these equations givesT ′′0 = 0T ′′2 + 4T 2 = 0T ′′3 + 9T 3 = cos 2tT 0 (0) = 1 2, T ′ 0(0) = 0T 2 (0) = 1 2, T ′ 2(0) = 0T 3 (0) = 0 , T ′ 3(0) = 0The final solution isT 0 (t) = 1 2T 2 (t) = 1 cos 2t2T 3 (t) = 1 (cos 2t − cos 3t)5u(x, t) = xt + 1 2 + 1 2 cos 2t cos 2x + 1 (cos 2t − cos 3t) cos 3x5(b) To make the boundary conditions homogeneous we write(u(x, t) = v(x, t) + x sin t + x 1 − x )2π9