We are asked to solve the heat equation [ u_t = u_{xx}, \quad 0 \le x \le 1, , t>0 ] with [ u(x,0) = 0, \quad u(0,t) = 0, \quad u(1,t) = t ] using the Crank–Nicolson method with (h = 0.2) and (\lambda = 1), for two time steps. --- Step 1: Discretization * Spatial step: (h = 0.2 \implies x_0 = 0, x_1 = 0.2, x_2 = 0.4, x_3 = 0.6, x_4 = 0.8, x_5 = 1) * Time step: let (k) satisfy (\lambda = \frac{k}{h^2} = 1 \implies k = h^2 = 0.04) * Interior nodes: (i = 1,2,3,4) (since (u_0) and (u_5) are boundary values) Crank–Nicolson formula for interior nodes: [ -\frac{\lambda}{2} u_{i-1}^{n+1} + (1+\lambda) u_i^{n+1} - \frac{\lambda}{2} u_{i+1}^{n+1} = \frac{\lambda}{2} u_{i-1}^n + (1-\lambda) u_i^n + \frac{\lambda}{2} u_{i+1}^n ] Given (\lambda = 1), this simplifies to: [ -0.5, u_{i-1}^{n+1} + 2, u_i^{n+1} - 0.5, u_{i+1}^{n+1} = 0.5, u_{i-1}^{n} + 0, u_i^n + 0.5, u_{i+1}^n ] --- Step 2: Initial and boundary conditions * (u_i^0 = 0) for all interior nodes * (u_0^n = 0), (u_5^n = t_n = n k) Step 3: Construct the system for (n=0 \to n=1) Interior nodes (u_1, u_2, u_3, u_4): Matrix (A) for LHS: [ A = \begin{bmatrix} 2 & -0.5 &