Suppose −u′′=2, with the boundary condition u(1)=0 changed to u′(1)=0. This ``natural'' condition on u′ need not be imposed on the trial functions V. With h=31, there is an extra half-hat V3, which goes from 0 to 1 between x=32 and x=1. Compute A33=∫(V3′)2dx and f3=∫2V3dx. Solve Ay=f for the finite element solution y1V1+y2V2+y3V3.