Let's solve both problems using **separation of variables**. --- ## **9.** \[ k \frac{\partial^2 u}{\partial x^2} - u = \frac{\partial u}{\partial t}, \quad k > 0 \] **Step 1: Assume a product solution** Let \( u(x, t) = X(x)T(t) \). **Step 2: Substitute into the PDE** \[ k \frac{\partial^2 u}{\partial x^2} - u = \frac{\partial u}{\partial t} \] \[ k X''(x)T(t) - X(x)T(t) = X(x)T'(t) \] **Step 3: Rearrange and separate variables** \[ k X''(x)T(t) - X(x)T(t) - X(x)T'(t) = 0 \] \[ k X''(x)T(t) - X(x)T(t) = X(x)T'(t) \] \[ k X''(x)T(t) - X(x)T(t) - X(x)T'(t) = 0 \] \[ k X''(x)T(t) - X(x)T(t) = X(x)T'(t) \] Divide both sides by \( X(x)T(t) \) (assuming neither is zero): \[ k \frac{X''(x)}{X(x)} - 1 = \frac{T'(t)}{T(t)} \] Set both sides equal to a constant, say \(-\lambda\): \[ k \frac{X''(x)}{X(x)} - 1 = \frac{T'(t)}{T(t)} = -\lambda \] --- ### **Spatial ODE** \[ k X''(x) + \lambda X(x) - X(x) = 0 \] \[ k X''(x) + (\lambda - 1) X(x) = 0 \] \[ X''(x) + \frac{\lambda - 1}{k} X(x) = 0 \] Let \(\alpha^2 = \frac{\lambda - 1}{k}\): \[ X''(x) + \alpha^2 X(x) = 0 \] General solution: \[ X(x) = A \cos(\alpha x) + B \sin(\alpha x) \] --- ### **Temporal ODE** \[ \frac{T'(t)}{T(t)} = -\lambda \implies T'(t) = -\lambda T(t) \] General solution: \[ T(t) = C e^{-\lambda t} \] --- ### **Final Product Solution** \[ u(x, t) = [A \cos(\alpha x) + B \sin(\alpha x)] \cdot C e^{-\lambda t} \] where \(\alpha^2 = \frac{\lambda - 1}{k}\). --- ## **10.** \[ k \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t}, \quad k > 0 \] This is the **heat equation**. **Step 1: Assume a product solution** Let \( u(x, t) = X(x)T(t) \). **Step 2: Substitute into the PDE** \[ k X''(x)T(t) = X(x)T'(t) \] Divide both sides by \( X(x)T(t) \): \[ k \frac{X''(x)}{X(x)} = \frac{T'(t)}{T(t)} = -\lambda \] --- ### **Spatial ODE** \[ X''(x) + \frac{\lambda}{k} X(x) = 0 \] Let \(\mu^2 = \frac{\lambda}{k}\): \[ X''(x) + \mu^2 X(x) = 0 \] General solution: \[ X(x) = D \cos(\mu x) + E \sin(\mu x) \] --- ### **Temporal ODE** \[ T'(t) + \lambda T(t) = 0 \] \[ T(t) = F e^{-\lambda t} \] --- ### **Final Product Solution** \[ u(x, t) = [D \cos(\mu x) + E \sin(\mu x)] \cdot F e^{-\lambda t} \] where \(\mu^2 = \frac{\lambda}{k}\). --- ## **Summary of Answers** ### **9.** \[ \boxed{ u(x, t) = [A \cos(\alpha x) + B \sin(\alpha x)] \, e^{-\lambda t}, \quad \alpha^2 = \frac{\lambda - 1}{k} } \] ### **10.** \[ \boxed{ u(x, t) = [D \cos(\mu x) + E \sin(\mu x)] \, e^{-\lambda t}, \quad \mu^2 = \frac{\lambda}{k} } \] These are the separated variables, product solutions for the given PDEs. Boundary and initial conditions would allow you to specify the constants and the eigenvalues.