Let's solve **problem 5** using separation of variables: ### **Problem 5** \[ x \frac{\partial u}{\partial x} = y \frac{\partial u}{\partial y} \] --- ## **Step 1: Assume a product solution** Let \( u(x, y) = X(x)Y(y) \). --- ## **Step 2: Compute the partial derivatives** \[ \frac{\partial u}{\partial x} = Y(y) \frac{dX}{dx}, \quad \frac{\partial u}{\partial y} = X(x) \frac{dY}{dy} \] --- ## **Step 3: Substitute into the PDE** \[ x \left[ Y(y) \frac{dX}{dx} \right] = y \left[ X(x) \frac{dY}{dy} \right] \] \[ x Y(y) X'(x) = y X(x) Y'(y) \] --- ## **Step 4: Rearrange and separate variables** Divide both sides by \( X(x)Y(y) \) (assuming neither is zero): \[ x \frac{X'(x)}{X(x)} = y \frac{Y'(y)}{Y(y)} \] --- ## **Step 5: Set both sides equal to a constant** Let \( x \frac{X'(x)}{X(x)} = y \frac{Y'(y)}{Y(y)} = \lambda \), where \( \lambda \) is a separation constant. So we have two ODEs: \[ x \frac{X'(x)}{X(x)} = \lambda \implies \frac{X'(x)}{X(x)} = \frac{\lambda}{x} \] \[ y \frac{Y'(y)}{Y(y)} = \lambda \implies \frac{Y'(y)}{Y(y)} = \frac{\lambda}{y} \] --- ## **Step 6: Solve the ODEs** **For \( X(x) \):** \[ \frac{X'(x)}{X(x)} = \frac{\lambda}{x} \implies \frac{dX}{X} = \lambda \frac{dx}{x} \implies \int \frac{dX}{X} = \lambda \int \frac{dx}{x} \implies \ln|X| = \lambda \ln|x| + C_1 \implies X(x) = C_1' x^{\lambda} \] **For \( Y(y) \):** \[ \frac{Y'(y)}{Y(y)} = \frac{\lambda}{y} \implies \frac{dY}{Y} = \lambda \frac{dy}{y} \implies \int \frac{dY}{Y} = \lambda \int \frac{dy}{y} \implies \ln|Y| = \lambda \ln|y| + C_2 \implies Y(y) = C_2' y^{\lambda} \] --- ## **Step 7: Write the general solution** \[ u(x, y) = X(x) Y(y) = (C_1' x^{\lambda})(C_2' y^{\lambda}) = C x^{\lambda} y^{\lambda} \] where \( C = C_1' C_2' \). Or, more generally, by superposition for all possible values of \( \lambda \): \[ \boxed{ u(x, y) = \sum_{\lambda} C_{\lambda} (x y)^{\lambda} } \] Or, in the simplest separated form: \[ \boxed{ u(x, y) = f(x y) } \] where \( f \) is any function that can be written as a sum of powers. --- ### **Final Answer (for Problem 5):** \[ \boxed{ u(x, y) = C (x y)^{\lambda} } \] or the general solution is any function of the product \( x y \): \[ \boxed{ u(x, y) = f(x y) } \] --- Would you like the step-by-step for any of the other problems?

Certainly! Please specify which other problem you'd like me to solve step-by-step. If you provide the problem statement or the PDE, I can proceed accordingly.

Let's proceed with **Problem 6** using separation of variables: --- ## **Problem 6** The PDE given is: \[ x^2 \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0 \] --- ## **Step 1: Assume a product solution** Suppose: \[ u(x, y) = X(x) Y(y) \] --- ## **Step 2: Compute the second derivatives** \[ \frac{\partial^2 u}{\partial x^2} = Y(y) X''(x) \] \[ \frac{\partial^2 u}{\partial y^2} = X(x) Y''(y) \] --- ## **Step 3: Substitute into the PDE** \[ x^2 \left[ Y(y) X''(x) \right] + y^2 \left[ X(x) Y''(y) \right] = 0 \] \[ x^2 Y(y) X''(x) + y^2 X(x) Y''(y) = 0 \] --- ## **Step 4: Divide by \( X(x) Y(y) \)** Assuming \( X(x) \neq 0 \), \( Y(y) \neq 0 \): \[ x^2 \frac{X''(x)}{X(x)} + y^2 \frac{Y''(y)}{Y(y)} = 0 \] --- ## **Step 5: Separate variables** Rewrite as: \[ x^2 \frac{X''(x)}{X(x)} = - y^2 \frac{Y''(y)}{Y(y)} = \lambda \] where \( \lambda \) is a separation constant. Thus, we get two ODEs: \[ x^2 \frac{X''(x)}{X(x)} = \lambda \] \[ y^2 \frac{Y''(y)}{Y(y)} = -\lambda \] --- ## **Step 6: Write the ODEs explicitly** **For \( X(x) \):** \[ x^2 X''(x) - \lambda X(x) = 0 \] \[ X''(x) - \frac{\lambda}{x^2} X(x) = 0 \] **For \( Y(y) \):** \[ y^2 Y''(y) + \lambda Y(y) = 0 \] \[ Y''(y) + \frac{\lambda}{y^2} Y(y) = 0 \] --- ## **Step 7: Recognize the form of the ODEs** These are Cauchy-Euler (or equidimensional) equations, similar to the standard forms: \[ x^2 X'' - \lambda X = 0 \] which suggests solutions involving power functions depending on the sign of \( \lambda \). --- ## **Step 8: Solve the \( X \)-equation** **Case 1: \( \lambda = m^2 \geq 0 \) (for real solutions)** The equation: \[ X''(x) - \frac{m^2}{x^2} X(x) = 0 \] has solutions: \[ X(x) = A x^{m} + B x^{-m} \] Similarly, for \( Y(y) \): \[ Y''(y) + \frac{m^2}{y^2} Y(y) = 0 \] which has solutions: \[ Y(y) = C y^{m} + D y^{-m} \] --- ## **Step 9: Write the general solution** Combining, the product solution: \[ u(x, y) = \left( A x^{m} + B x^{-m} \right) \left( C y^{m} + D y^{-m} \right) \] or, more generally, superpositions over all \( m \): \[ u(x, y) = \sum_{m} \left[ A_m x^{m} + B_m x^{-m} \right] \left[ C_m y^{m} + D_m y^{-m} \right] \] --- ## **Final solution:** \[ \boxed{ u(x, y) = \sum_{m} \left[ a_m x^{m} + b_m x^{-m} \right] \left[ c_m y^{m} + d_m y^{-m} \right] } \] where \( m \) is a parameter (possibly continuous or discrete depending on boundary conditions), and the coefficients are determined by boundary conditions. --- **Summary:** - The PDE separates into ODEs with solutions involving power functions. - The general solution is a superposition of products of powers of \( x \) and \( y \). --- Would you like me to clarify any step further or assist with another problem?

Thank you for the clarification. The correct PDE for **Problem 6** is: \[ y \frac{\partial u}{\partial x} = x \frac{\partial u}{\partial y} \] Let's solve this problem step-by-step using separation of variables: --- ## **Problem 6 (corrected):** \[ y \frac{\partial u}{\partial x} = x \frac{\partial u}{\partial y} \] --- ## **Step 1: Assume a product solution** Assume: \[ u(x, y) = X(x) Y(y) \] --- ## **Step 2: Compute partial derivatives** \[ \frac{\partial u}{\partial x} = Y(y) X'(x) \] \[ \frac{\partial u}{\partial y} = X(x) Y'(y) \] --- ## **Step 3: Substitute into the PDE** \[ y \left[ Y(y) X'(x) \right] = x \left[ X(x) Y'(y) \right] \] \[ y Y(y) X'(x) = x X(x) Y'(y) \] --- ## **Step 4: Divide both sides by \( X(x) Y(y) \)** \[ \frac{y Y(y) X'(x)}{X(x) Y(y)} = \frac{x X(x) Y'(y)}{X(x) Y(y)} \] \[ \frac{y X'(x)}{X(x)} = \frac{x Y'(y)}{Y(y)} \] --- ## **Step 5: Set both sides equal to a constant** Let: \[ \frac{y X'(x)}{X(x)} = \frac{x Y'(y)}{Y(y)} = \lambda \] where \( \lambda \) is a separation constant. This gives two equations: \[ \frac{X'(x)}{X(x)} = \frac{\lambda}{y} \] \[ \frac{Y'(y)}{Y(y)} = \frac{\lambda}{x} \] --- ## **Step 6: Express derivatives** From the first: \[ X'(x) = \frac{\lambda}{y} X(x) \] From the second: \[ Y'(y) = \frac{\lambda}{x} Y(y) \] But note that these involve both variables; to proceed, observe that the original structure suggests a different approach: **change of variables**. --- ## **Alternative approach: change of variables** Notice that the PDE relates derivatives with coefficients \( y \) and \( x \), symmetric in a certain way; it suggests that the solution depends on the ratio \( x/y \). --- ## **Step 7: Recognize an invariance** Suppose \( u \) depends on the ratio \( t = \frac{x}{y} \). Let's define: \[ u(x, y) = f(t) \quad \text{where} \quad t = \frac{x}{y} \] --- ## **Step 8: Compute derivatives in terms of \( t \)** \[ u = f(t) \] \[ \frac{\partial u}{\partial x} = f'(t) \frac{\partial t}{\partial x} = f'(t) \cdot \frac{1}{y} \] \[ \frac{\partial u}{\partial y} = f'(t) \frac{\partial t}{\partial y} = f'(t) \cdot \left( - \frac{x}{y^2} \right) = - f'(t) \cdot \frac{x}{y^2} \] --- ## **Step 9: Substitute into the PDE** \[ y \cdot \frac{\partial u}{\partial x} = x \cdot \frac{\partial u}{\partial y} \] becomes: \[ y \cdot \left( \frac{f'(t)}{y} \right) = x \cdot \left( - \frac{f'(t) x}{y^2} \right) \] Simplify: \[ f'(t) = - \frac{f'(t) x^2}{y^2} \] Recall \( t = \frac{x}{y} \), so \( x = t y \). Substitute: \[ f'(t) = - \frac{f'(t) (t y)^2}{y^2} = - f'(t) t^2 \] So: \[ f'(t) = - t^2 f'(t) \] --- ## **Step 10: Solve the resulting ODE** Rearranged: \[ f'(t) + t^2 f'(t) = 0 \implies f'(t)(1 + t^2) = 0 \] Thus: \[ f'(t) = 0 \] which implies: \[ f(t) = \text{constant} \] --- ## **## **Final solution:** \[ u(x, y) = \text{constant} \] --- ## **Summary:** - The general solution to the PDE: \[ y \frac{\partial u}{\partial x} = x \frac{\partial u}{\partial y} \] is **any constant function**. --- ## **Alternative interpretation:** - Since the derivatives lead to \( f'(t) = 0 \), the only solutions are constants. - Alternatively, the solution is **any function of \( x/y \)** with a specific form, but from the derivation, the only solutions are constants. --- **If boundary conditions are given, the particular solution might differ.** --- Would you like me to explore other solution methods or clarify further?

Let's carefully analyze and solve **Problem 7**, which involves the PDE: \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} = 0 \] --- ## Step 1: Recognize the type of PDE This is a second-order linear PDE with mixed derivatives. To analyze its nature, we consider the **second-order PDE classification** via the **discriminant**: \[ A = 1, \quad B = 1, \quad C = 1 \] The discriminant: \[ D = B^2 - 4AC = 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3 < 0 \] Since \( D < 0 \), the PDE is **elliptic**. --- ## Step 2: Assume a solution via separation of variables Suppose: \[ u(x, y) = X(x) Y(y) \] --- ## Step 3: Compute second derivatives \[ \frac{\partial^2 u}{\partial x^2} = X''(x) Y(y) \] \[ \frac{\partial^2 u}{\partial y^2} = X(x) Y''(y) \] \[ \frac{\partial^2 u}{\partial x \partial y} = X'(x) Y'(y) \] --- ## Step 4: Substitute into the PDE \[ X''(x) Y(y) + X'(x) Y'(y) + X(x) Y''(y) = 0 \] Divide through by \( X(x) Y(y) \): \[ \frac{X''(x)}{X(x)} + \frac{X'(x)}{X(x)} \frac{Y'(y)}{Y(y)} + \frac{Y''(y)}{Y(y)} = 0 \] This expression involves mixed derivatives, which complicates variables separation directly in the standard way. Alternatively, **try a change of variables** to diagonalize the PDE. --- ## Step 5: Change of variables to simplify the PDE The PDE's second-order part can be written as: \[ u_{xx} + u_{xy} + u_{yy} = 0 \] Expressed in matrix form: \[ \begin{bmatrix} u_{xx} & u_{xy} \\ u_{yx} & u_{yy} \end{bmatrix} \] with quadratic form: \[ A \, \mathrm{d}x^2 + 2B \, \mathrm{d}x \mathrm{d}y + C \, \mathrm{d}y^2 \] which in our case is: \[ \mathrm{d}s^2 = dx^2 + 2 dx dy + dy^2 \] Corresponds to the quadratic form: \[ Q(\mathrm{d}x, \mathrm{d}y) = dx^2 + 2 dx dy + dy^2 \] --- ## Step 6: Diagonalize the quadratic form The matrix associated: \[ \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix} \] Eigenvalues: \[ \lambda = 1 \pm \frac{1}{2} \] Calculating: \[ \det \begin{bmatrix} 1 - \lambda & \frac{1}{2} \\ \frac{1}{2} & 1 - \lambda \end{bmatrix} = (1 - \lambda)^2 - \frac{1}{4} = 0 \] \[ (1 - \lambda)^2 = \frac{1}{4} \] \[ 1 - \lambda = \pm \frac{1}{2} \] \[ \lambda_1 = 1 - \frac{1}{2} = \frac{1}{2} \] \[ \lambda_2 = 1 + \frac{1}{2} = \frac{3}{2} \] Corresponding eigenvectors lead to new variables \( \xi, \eta \) that diagonalize the form. --- ## **Step 7: Find new variables** Let us define: \[ \xi = x + y \quad \text{(from the eigenvector corresponding to } \lambda_1) \] \[ \eta = x - y \quad \text{(from the eigenvector corresponding to } \lambda_2) \] Transforming the PDE into these variables (details involve linear algebra and chain rule), the PDE becomes **a second-order elliptic PDE** in the new variables. --- ## **Step 8: Attempt a solution in the transformed variables** Alternatively, given the nature of the PDE, **try a solution of the form**: \[ u(x, y) = e^{\alpha x + \beta y} \] --- ## Step 9: Plug in the exponential solution Compute derivatives: \[ u_x = \alpha e^{\alpha x + \beta y} \] \[ u_{xx} = \alpha^2 e^{\alpha x + \beta y} \] \[ u_y = \beta e^{\alpha x + \beta y} \] \[ u_{yy} = \beta^2 e^{\alpha x + \beta y} \] \[ u_{xy} = \alpha \beta e^{\alpha x + \beta y} \] Substitute into the PDE: \[ \alpha^2 e^{\alpha x + \beta y} + \alpha \beta e^{\alpha x + \beta y} + \beta^2 e^{\alpha x + \beta y} = 0 \] Dividing through: \[ \alpha^2 + \alpha \beta + \beta^2 = 0 \] --- ## **Step 10: Solve the characteristic equation** \[ \alpha^2 + \alpha \beta + \beta^2 = 0 \] Let \( r = \frac{\alpha}{\beta} \) (assuming \( \beta \neq 0 \)): \[ r^2 + r + 1 = 0 \] Discriminant: \[ \Delta = 1 - 4 = -3 < 0 \] Solutions: \[ r = \frac{-1 \pm i \sqrt{3}}{2} \] Thus, \[ \frac{\alpha}{\beta} = r = \frac{-1 \pm i \sqrt{3}}{2} \] which implies: \[ \alpha = r \beta \] Since the roots are complex conjugates, the general solution involves **complex exponentials**, leading to real solutions via Euler's formula. --- ## **Final general solution:** The general solution can be expressed as a linear combination of exponential functions: \[ u(x, y) = e^{\alpha x + \beta y} \] where \( (\alpha, \beta) \) satisfy: \[ \alpha^2 + \alpha \beta + \beta^2 = 0 \] or equivalently, the general real solution is: \[ u(x, y) = \text{Re} \left\{ e^{r \beta x + \beta y} \right\} \] for the roots \( r = \frac{-1 \pm i \sqrt{3}}{2} \). --- ## **Alternative: Polynomial or trigonometric solutions** Given the elliptic nature, solutions may also be expressed via harmonic functions, but the exponential form with complex roots captures the general solution structure. --- ## **Summary:** - The PDE is elliptic. - The exponential solutions lead to a characteristic equation with complex roots. - The general solution involves linear combinations of functions like: \[ u(x, y) = e^{a x} \cos(b y) \quad \text{and} \quad e^{a x} \sin(b y) \] where \( a \) and \( b \) are related via the roots. --- If you require explicit forms or boundary conditions, please specify!

Let's analyze and solve **Problem 8** step-by-step. --- ## **Problem 8:** Suppose the PDE is: \[ x^2 \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0 \] *(Note: if the actual problem differs, please specify the exact PDE. For now, I proceed with this common form.)* --- ### **Step 1: Recognize the type of PDE** The PDE is: \[ x^2 u_{xx} + y^2 u_{yy} = 0 \] which is a **second-order linear PDE** with variable coefficients. It resembles a form of **elliptic/hyperbolic** PDE depending on the coefficients. --- ### **Step 2: Attempt separation of variables** Assuming a solution of the form: \[ u(x, y) = X(x) Y(y) \] which leads to derivatives: \[ u_{xx} = X''(x) Y(y), \quad u_{yy} = X(x) Y''(y) \] Substituting into the PDE: \[ x^2 X''(x) Y(y) + y^2 X(x) Y''(y) = 0 \] Divide through by \( X(x) Y(y) \): \[ \frac{x^2 X''(x)}{X(x)} + \frac{y^2 Y''(y)}{Y(y)} = 0 \] --- ### **Step 3: Separate variables** Set: \[ \frac{x^2 X''(x)}{X(x)} = - \lambda, \quad \frac{y^2 Y''(y)}{Y(y)} = \lambda \] where \( \lambda \) is a separation constant. This gives two ODEs: \[ x^2 X''(x) + \lambda X(x) = 0 \] \[ y^2 Y''(y) - \lambda Y(y) = 0 \] --- ### **Step 4: Solve the \( X \)-equation** \[ x^2 X'' + \lambda X = 0 \] This is a **Cauchy-Euler** type equation. The standard approach is to look for power solutions: \[ X(x) = x^{m} \] Compute derivatives: \[ X' = m x^{m-1} \] \[ X'' = m (m-1) x^{m-2} \] Substitute into the ODE: \[ x^2 \times m (m-1) x^{m-2} + \lambda x^{m} = 0 \] Simplify: \[ m (m-1) x^{m} + \lambda x^{m} = 0 \] \[ \left[ m (m-1) + \lambda \right] x^{m} = 0 \] Since \( x^{m} \neq 0 \), the **characteristic equation** is: \[ m (m - 1) + \lambda = 0 \] which yields: \[ m^2 - m + \lambda = 0 \] --- ### **Step 5: Solve for \( m \)** \[ m = \frac{1 \pm \sqrt{1 - 4 \lambda}}{2} \] --- ### **Step 6: Solve the \( Y \)-equation** \[ y^2 Y'' - \lambda Y = 0 \] Assuming \( Y(y) = y^{n} \): \[ Y' = n y^{n-1} \] \[ Y'' = n (n-1) y^{n-2} \] Substitute into the ODE: \[ y^2 \times n (n-1) y^{n-2} - \lambda y^{n} = 0 \] \[ n (n-1) y^{n} - \lambda y^{n} = 0 \] \[ \left[ n(n-1) - \lambda \right] y^{n} = 0 \] Thus: \[ n(n-1) = \lambda \] --- ### **Step 7: Express solutions** - For the \( X \)-equation: \[ m = \frac{1 \pm \sqrt{1 - 4 \lambda}}{2} \] - For the \( Y \)-equation: \[ n \text{ satisfies } n(n-1) = \lambda \] --- ### **Step 8: Compatibility condition** From the relations, the eigenvalues \( \lambda \) satisfy both: \[ n(n-1) = \lambda \] \[ m^2 - m + \lambda = 0 \] Substitute \( \lambda = n(n-1) \): \[ m^2 - m + n(n-1) = 0 \] which relates \( m \) and \( n \). The general solution involves arbitrary constants and superpositions over these solutions. --- ### **### Final general solution:** \[ u(x, y) = \sum_{i} \left[ A_i x^{m_i} \right] \left[ B_i y^{n_i} \right] \] where \( m_i \) and \( n_i \) satisfy the algebraic relations above. --- ## **Summary:** - The PDE admits solutions involving power functions \( x^{m} \) and \( y^{n} \). - The exponents satisfy quadratic equations related via the separation constant \( \lambda \). - The general solution is a combination (superposition) of such product solutions. --- If boundary conditions are available, they will determine the specific coefficients and allowed values of \( m, n \). --- **Would you like me to clarify any step further or assist with particular boundary conditions?**

Let's analyze and solve the PDE: \[ y \frac{\partial^2 u}{\partial x \partial y} + u = 0 \] --- ## Step 1: Recognize the type of PDE This is a second-order linear PDE with mixed derivatives. To understand its nature, note that the highest-order term involves \(\frac{\partial^2 u}{\partial x \partial y}\), multiplied by \( y \), plus a zeroth-order term \( u \). --- ## Step 2: Consider a solution of separable form Suppose: \[ u(x, y) = X(x) V(y) \] Alternatively, since the PDE involves mixed derivatives, it might be advantageous to look for solutions based on the structure of the equation. --- ## Step 3: Use the **method of separation of variables** Calculate the mixed derivative: \[ \frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial y} \left( \frac{\partial u}{\partial x} \right) \] Assuming \( u(x, y) = X(x) V(y) \), \[ \frac{\partial u}{\partial x} = X'(x) V(y) \] then, \[ \frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial y} \left( X'(x) V(y) \right) = X'(x) V'(y) \] Now, substitute into the PDE: \[ y \cdot X'(x) V'(y) + X(x) V(y) = 0 \] --- ## Step 4: Divide through by \( X(x) V(y) \) (assuming non-zero functions): \[ \frac{y X'(x)}{X(x)} \cdot \frac{V'(y)}{V(y)} + 1 = 0 \] Rearranged: \[ \frac{y X'(x)}{X(x)} \cdot \frac{V'(y)}{V(y)} = -1 \] --- ## Step 5: Separate variables Let: \[ \frac{y X'(x)}{X(x)} = \lambda \] and \[ \frac{V'(y)}{V(y)} = \frac{-1}{\lambda} \] where \(\lambda\) is a separation constant (non-zero, to avoid division by zero). --- ## Step 6: Derive ODEs for \( X(x) \) and \( V(y) \) From the definitions: \[ \frac{X'(x)}{X(x)} = \frac{\lambda}{y} \] but note that \( y \) is a variable, so for the separation to hold, \( \lambda \) must be zero or the approach needs to be refined. Alternatively, notice that the previous step indicates the product \( y \cdot X'(x) / X(x) \) is a constant, which depends on \( y \). To make progress, consider fixing the approach differently. --- ## Alternative approach: **try exponential solutions** Suppose: \[ u(x,y) = e^{k x} \phi(y) \] Compute derivatives: \[ \frac{\partial^2 u}{\partial x \partial y} = \frac{\partial}{\partial y} \left( k e^{k x} \phi(y) \right) = k e^{k x} \phi'(y) \] Substitute into PDE: \[ y \cdot k e^{k x} \phi'(y) + e^{k x} \phi(y) = 0 \] Divide both sides by \( e^{k x} \): \[ k y \phi'(y) + \phi(y) = 0 \] This is an ODE in \( y \): \[ k y \phi'(y) + \phi(y) = 0 \] --- ## Step 7: Solve the ODE for \( \phi(y) \) Rewrite: \[ k y \frac{d \phi}{dy} + \phi = 0 \] Divide through by \( y \): \[ k \frac{d \phi}{dy} + \frac{\phi}{y} = 0 \] or: \[ k \frac{d \phi}{dy} = - \frac{\phi}{y} \] Separate variables: \[ \frac{d \phi}{\phi} = - \frac{1}{k y} dy \] Integrate: \[ \ln |\phi| = - \frac{1}{k} \ln |y| + C \] \[ \phi(y) = C' y^{-1/k} \] --- ## **Step 8: Write the general solution** Recall the exponential ansatz: \[ u(x, y) = e^{k x} \phi(y) = C' e^{k x} y^{-1/k} \] Since \( k \) is arbitrary, the general solution is an integral superposition over all \( k \): \[ \boxed{ u(x, y) = \int_{-\infty}^\infty A(k) e^{k x} y^{-1/k} dk } \] where \( A(k) \) is an arbitrary function determined by boundary conditions. --- ## **Summary:** - The solutions are of the form: \[ u(x, y) = e^{k x} y^{-1/k} \] - The general solution is a superposition (integral) over all \( k \): \[ u(x, y) = \int_{-\infty}^\infty A(k) e^{k x} y^{-1/k} dk \] --- ## **Note:** - The structure suggests solutions involving exponential functions in \( x \) and power functions in \( y \). - Boundary conditions would specify \( A(k) \). --- If you'd like, I can help with particular boundary conditions or further analysis!