In class, we looked at the following boundary value problem du = x, 0 < x <1 dx u(0) = u(1) = 0. (a) Given that the Galerkin method produces a diagonal stiffness matrix for this problem with the trigonometric basis functions used in class, show that the n-th solution coefficient is given by the general expression 2(-1)+1 an (Hint: Start by computing the n-th diagonal entry for the stiffness matrix, Knn, and the n-th component of the load vector, Fn, both of which involve the n-th basis function On = sin(nx).) (b) Apply the Galerkin method using the polynomial basis functions of Problem 1, part d with N = 2 for this problem. Do you notice anything special about uN in this case? (c) Explain why the polynomial basis } 0 = 1, 02 = x, 03 = x,..., ON = xN could not be used as a basis for this problem (or Problem 1).