Let's solve **part (a)** step by step using the method of eigenfunction expansion. --- ### **Step 1: Homogenize the Boundary Conditions** Given: \[ u_t - k u_{xx} = x, \quad 0 \leq x \leq \pi \] with \[ u(0, t) = 1, \quad u(\pi, t) = 0, \quad u(x, 0) = 0 \] The boundary conditions are **not homogeneous**. Let's split \( u(x, t) \) as: \[ u(x, t) = v(x, t) + w(x) \] where \( v(x, t) \) satisfies homogeneous boundary conditions, and \( w(x) \) is a steady-state function chosen so that \( u(0, t) = 1 \), \( u(\pi, t) = 0 \). Let \( w(x) \) be a linear function: \[ w(x) = 1 - \frac{x}{\pi} \] so \( w(0) = 1 \), \( w(\pi) = 0 \). Now, \( v(x, t) \) satisfies: \[ v(0, t) = 0, \quad v(\pi, t) = 0 \] Substitute into the PDE: \[ u_t - k u_{xx} = x \implies v_t - k v_{xx} = x - k w''(x) \] But \( w''(x) = 0 \), so the equation is: \[ v_t - k v_{xx} = x \] Initial condition: \[ u(x, 0) = 0 \implies v(x, 0) = -w(x) = -\left(1 - \frac{x}{\pi}\right) \] --- ### **Step 2: Expand in Eigenfunctions** The homogeneous BCs suggest using sine series: \[ v(x, t) = \sum_{n=1}^\infty A_n(t) \sin(nx) \] Plug into the PDE: \[ \sum_{n=1}^\infty A_n'(t) \sin(nx) - k \sum_{n=1}^\infty A_n(t) n^2 \sin(nx) = x \] Expand \( x \) in sine series on \( [0, \pi] \): \[ x = \sum_{n=1}^\infty b_n \sin(nx) \] where \[ b_n = \frac{2}{\pi} \int_0^\pi x \sin(nx) dx \] Integrate by parts: Let \( u = x \), \( dv = \sin(nx) dx \), \( du = dx \), \( v = -\frac{1}{n}\cos(nx) \): \[ b_n = \frac{2}{\pi} \left[ -\frac{x}{n} \cos(nx) \Big|_0^\pi + \frac{1}{n} \int_0^\pi \cos(nx) dx \right] \] \[ = \frac{2}{\pi} \left( -\frac{\pi}{n} \cos(n\pi) + 0 + \frac{1}{n} \left[ \frac{1}{n} \sin(nx) \Big|_0^\pi \right] \right) \] \[ = \frac{2}{\pi} \left( -\frac{\pi}{n} (-1)^n + \frac{1}{n^2} (0 - 0) \right) = \frac{2}{\pi} \left( \frac{\pi}{n} (-1)^{n+1} \right) = \frac{2(-1)^{n+1}}{n} \] Now, the ODE for each \( A_n(t) \) is: \[ A_n'(t) + k n^2 A_n(t) = b_n \] This is a first-order linear ODE. The integrating factor is \( e^{k n^2 t} \): \[ \frac{d}{dt}\left(A_n(t) e^{k n^2 t}\right) = b_n e^{k n^2 t} \] Integrate both sides: \[ A_n(t) e^{k n^2 t} = b_n \int_0^t e^{k n^2 s} ds + C \] \[ A_n(t) = b_n e^{-k n^2 t} \int_0^t e^{k n^2 s} ds + C e^{-k n^2 t} \] \[ = b_n \frac{1 - e^{-k n^2 t}}{k n^2} + C e^{-k n^2 t} \] Use the initial condition \( v(x, 0) = -\left(1 - \frac{x}{\pi}\right) \). Expand \( 1 - \frac{x}{\pi} \) in sine series: \[ 1 - \frac{x}{\pi} = \sum_{n=1}^\infty d_n \sin(nx) \] where \[ d_n = \frac{2}{\pi} \int_0^\pi \left(1 - \frac{x}{\pi}\right) \sin(nx) dx \] \[ = \frac{2}{\pi} \left( \int_0^\pi \sin(nx) dx - \frac{1}{\pi} \int_0^\pi x \sin(nx) dx \right) \] \[ \int_0^\pi \sin(nx) dx = \left. -\frac{1}{n} \cos(nx) \right|_0^\pi = -\frac{1}{n} (\cos(n\pi) - 1) = \frac{1}{n} (1 - (-1)^n) \] So for odd \( n \), \( 1 - (-1)^n = 2 \), for even \( n \), 0. We already computed \( \int_0^\pi x \sin(nx) dx = \frac{\pi^2}{n} (-1)^{n+1} \). So, \[ d_n = \frac{2}{\pi} \left[ \frac{1}{n} (1 - (-1)^n) - \frac{1}{\pi} \frac{\pi^2}{n} (-1)^{n+1} \right] \] \[ = \frac{2}{\pi} \left[ \frac{1}{n} (1 - (-1)^n) - \frac{\pi}{n} (-1)^{n+1} \right] \] But you can leave it as \( d_n \). So initial condition for \( A_n(0) \): \[ A_n(0) = -d_n \] So, \[ A_n(t) = b_n \frac{1 - e^{-k n^2 t}}{k n^2} - d_n e^{-k n^2 t} \] --- ### **Final Answer** \[ \boxed{ u(x, t) = 1 - \frac{x}{\pi} + \sum_{n=1}^\infty \left[ \frac{2(-1)^{n+1}}{n k n^2} (1 - e^{-k n^2 t}) - d_n e^{-k n^2 t} \right] \sin(nx) } \] where \[ d_n = \frac{2}{\pi} \left[ \frac{1}{n} (1 - (-1)^n) - \frac{\pi}{n} (-1)^{n+1} \right] \] --- **Summary of Steps:** 1. **Homogenize the boundary conditions** by splitting \( u(x, t) \) into a steady-state part and a part with homogeneous BCs. 2. **Expand the solution in eigenfunctions** (sine series), solve the resulting ODEs for the coefficients, and apply initial conditions to find the full solution.