Let's break the problem into 3 steps, as requested, with explanations after each step. --- ### **Step 1: Set up the Separation of Variables** **Equation:** \[ \nabla^2 V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = 0 \] Assume a solution of the form: \[ V(x, y) = X(x)Y(y) \] Plug into the Laplace equation: \[ \frac{\partial^2}{\partial x^2}[X(x)Y(y)] + \frac{\partial^2}{\partial y^2}[X(x)Y(y)] = 0 \] \[ Y(y) X''(x) + X(x) Y''(y) = 0 \] Divide both sides by \(X(x) Y(y)\): \[ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 \] Set each term equal to a separation constant \(-k^2\) (as suggested): \[ \frac{X''(x)}{X(x)} = -k^2, \quad \frac{Y''(y)}{Y(y)} = k^2 \] **Explanation:** We have separated the variables and set up two ordinary differential equations (ODEs) for \(X(x)\) and \(Y(y)\). --- ### **Step 2: Solve the ODEs for \(X(x)\) and \(Y(y)\)** **For \(X(x)\):** \[ X''(x) + k^2 X(x) = 0 \] General solution: \[ X(x) = A \cos(kx) + B \sin(kx) \] **For \(Y(y)\):** \[ Y''(y) - k^2 Y(y) = 0 \] General solution: \[ Y(y) = C e^{k y} + D e^{-k y} \] **Explanation:** We've solved both ODEs, finding the general forms for \(X(x)\) and \(Y(y)\). The total solution can be written as a sum over all possible \(k\): \[ V(x, y) = \sum_k [A_k \cos(kx) + B_k \sin(kx)] [C_k e^{k y} + D_k e^{-k y}] \] --- ### **Step 3: Apply Boundary Conditions and Find the Specific Solution** **Boundary Conditions:** - \(V(x, y) \to 0\) as \(y \to \infty\) - \(V(0, y) = V(a, y) = 0\) - \(V(x, 0) = V_0\) (a constant) **Apply conditions:** 1. **As \(y \to \infty\), \(V \to 0\):** This requires \(C_k = 0\) (so the term with \(e^{k y}\) vanishes). 2. **\(V(0, y) = 0:\)** At \(x = 0\), \(V(0, y) = A_k \cdot 1 + B_k \cdot 0 = 0\) ⇒ \(A_k = 0\). 3. **\(V(a, y) = 0:\)** At \(x = a\), \(V(a, y) = B_k \sin(k a) D_k e^{-k y} = 0\). For non-trivial solution, \(\sin(k a) = 0\) ⇒ \(k = n\pi/a\), \(n=1,2,3,\ldots\). So, \(V(x, y) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right) e^{-n\pi y/a}\). 4. **\(V(x, 0) = V_0\):** \[ V(x, 0) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right) = V_0 \] Expand \(V_0\) as a Fourier sine series: \[ V_0 = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{a}\right) \] The coefficients are: \[ B_n = \frac{2}{a} \int_0^a V_0 \sin\left(\frac{n\pi x}{a}\right) dx = \frac{2 V_0}{a} \left[ -\frac{a}{n\pi} \cos\left(\frac{n\pi x}{a}\right) \right]_0^a \] \[ = \frac{2 V_0}{n\pi} (1 - (-1)^n) \] Only odd \(n\) gives non-zero coefficients (\(n=1,3,5,...\)). **Explanation:** We applied the boundary conditions, resulting in a Fourier sine series expansion, and found the specific form of the solution. --- ## **Final Answer** \[ \boxed{ V(x, y) = \sum_{n=1,3,5,\ldots}^{\infty} \frac{4 V_0}{n\pi} \sin\left(\frac{n\pi x}{a}\right) e^{-n\pi y/a} } \] where \(n\) runs over odd integers. This satisfies all the given boundary conditions.