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The finite difference scheme also gives exact solution at the nodes to the problem $-u''=1$ because $$\frac{u_{j-1} - 2 u_j + u_{j+1}}{h^2}$$ is exact for a quadratic function. For the more general case $$-u''(x) = f(x)$$ if you compute the integral $(f,\phi_i)$ exactly in $P_1$ FEM, it gives exact solution at the nodes. The FD scheme will not be exact ...