Let's solve **Problem 1** step-by-step using **separation of variables**: --- ### **Problem 1** \[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} \] --- #### **Step 1: Assume a Product Solution** Let \( u(x, y) = X(x)Y(y) \). --- #### **Step 2: Substitute Into the PDE** \[ \frac{\partial u}{\partial x} = Y(y) \frac{dX}{dx} \] \[ \frac{\partial u}{\partial y} = X(x) \frac{dY}{dy} \] Substitute into the equation: \[ Y(y) \frac{dX}{dx} = X(x) \frac{dY}{dy} \] --- #### **Step 3: Rearrange and Separate Variables** Divide both sides by \( X(x)Y(y) \): \[ \frac{1}{X(x)} \frac{dX}{dx} = \frac{1}{Y(y)} \frac{dY}{dy} \] Let both sides equal a constant, say \(\lambda\): \[ \frac{1}{X(x)} \frac{dX}{dx} = \lambda \] \[ \frac{1}{Y(y)} \frac{dY}{dy} = \lambda \] --- #### **Step 4: Solve the ODEs** **For \(X(x)\):** \[ \frac{dX}{dx} = \lambda X(x) \implies X(x) = A e^{\lambda x} \] **For \(Y(y)\):** \[ \frac{dY}{dy} = \lambda Y(y) \implies Y(y) = B e^{\lambda y} \] --- #### **Step 5: Write the General Solution** \[ u(x, y) = X(x) Y(y) = A e^{\lambda x} \cdot B e^{\lambda y} = C e^{\lambda (x + y)} \] where \(C = AB\). --- ### **Final Answer:** \[ \boxed{ u(x, y) = C e^{\lambda (x + y)} } \] where \(C\) and \(\lambda\) are arbitrary constants. --- Would you like to see the solution for the next problem as well?

Certainly! Let's proceed with **Problem 2** step-by-step using **separation of variables**. --- ### **Problem 2** \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \] --- #### **Step 1: Assume a Product Solution** Let \( u(x, y) = X(x)Y(y) \). --- #### **Step 2: Compute Partial Derivatives** \[ \frac{\partial u}{\partial x} = Y(y) \frac{dX}{dx} \] \[ \frac{\partial u}{\partial y} = X(x) \frac{dY}{dy} \] Substitute into the PDE: \[ x Y(y) \frac{dX}{dx} + y X(x) \frac{dY}{dy} = 0 \] --- #### **Step 3: Divide and Separate Variables** Divide through by \( X(x)Y(y) \): \[ x \frac{1}{X} \frac{dX}{dx} + y \frac{1}{Y} \frac{dY}{dy} = 0 \] Rewrite: \[ x \frac{1}{X} \frac{dX}{dx} = - y \frac{1}{Y} \frac{dY}{dy} \] Notice the left depends only on \(x\), and the right only on \(y\), so both must equal a constant, say \(\lambda\): \[ x \frac{1}{X} \frac{dX}{dx} = \lambda \] \[ - y \frac{1}{Y} \frac{dY}{dy} = \lambda \] --- #### **Step 4: Solve the ODEs** **For \(X(x)\):** \[ x \frac{dX}{dx} = \lambda X \] \[ \frac{dX}{dx} = \frac{\lambda}{x} X \] This is a separable differential equation: \[ \frac{1}{X} dX = \frac{\lambda}{x} dx \] Integrate both sides: \[ \ln |X| = \lambda \ln |x| + C_1 \] \[ X(x) = A x^{\lambda} \] **For \(Y(y)\):** \[ - y \frac{dY}{dy} = \lambda Y \] \[ \frac{dY}{dy} = - \frac{\lambda}{y} Y \] Similarly: \[ \frac{1}{Y} dY = - \frac{\lambda}{y} dy \] Integrate: \[ \ln |Y| = - \lambda \ln |y| + C_2 \] \[ Y(y) = B y^{-\lambda} \] --- ### **Step 5: Write the General Solution** \[ u(x, y) = X(x) Y(y) = A x^{\lambda} \cdot B y^{-\lambda} = C x^{\lambda} y^{-\lambda} \] where \( C = AB \). --- ### **Final Answer:** \[ \boxed{ u(x, y) = C \left( \frac{x}{y} \right)^{\lambda} } \] where \( C \) and \( \lambda \) are arbitrary constants. --- Would you like to continue with the next problem?

Let's carefully analyze **Problem 2** to identify any potential issues. --- ### **Problem 2** \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \] --- ### **Step 1: Assume a Product Solution** Suppose \( u(x, y) = X(x) Y(y) \). --- ### **Step 2: Compute Partial Derivatives** \[ \frac{\partial u}{\partial x} = Y(y) \frac{dX}{dx} \] \[ \frac{\partial u}{\partial y} = X(x) \frac{dY}{dy} \] Substitute into the PDE: \[ x Y(y) \frac{dX}{dx} + y X(x) \frac{dY}{dy} = 0 \] --- ### **Step 3: Divide and Separate Variables** Divide through by \( X(x) Y(y) \): \[ x \frac{1}{X} \frac{dX}{dx} + y \frac{1}{Y} \frac{dY}{dy} = 0 \] This can be written as: \[ x \frac{d}{dx} \ln X(x) + y \frac{d}{dy} \ln Y(y) = 0 \] --- ### **Step 4: Recognize the Structure** Notice that the first term depends only on \(x\), the second only on \(y\), and the sum equals zero: \[ x \frac{d}{dx} \ln X(x) = - y \frac{d}{dy} \ln Y(y) \] Since the left depends only on \(x\) and the right only on \(y\), both must be equal to a constant, say \(\lambda\): \[ x \frac{d}{dx} \ln X(x) = \lambda \] \[ - y \frac{d}{dy} \ln Y(y) = \lambda \] --- ### **Step 5: Solve the resulting ODEs** **For \(X(x)\):** \[ x \frac{d}{dx} \ln X(x) = \lambda \implies \frac{d}{dx} \ln X(x) = \frac{\lambda}{x} \] Integrate: \[ \ln X(x) = \lambda \ln x + C_1 \] \[ X(x) = A x^{\lambda} \] **For \(Y(y)\):** \[ - y \frac{d}{dy} \ln Y(y) = \lambda \implies \frac{d}{dy} \ln Y(y) = - \frac{\lambda}{y} \] Integrate: \[ \ln Y(y) = - \lambda \ln y + C_2 \] \[ Y(y) = B y^{-\lambda} \] --- ### **Step 6: Final Solution** \[ u(x, y) = X(x) Y(y) = A x^{\lambda} \cdot B y^{-\lambda} = C \left( \frac{x}{y} \right)^{\lambda} \] where \( C = AB \). --- ### **Conclusion:** This derivation is **correct** mathematically. The solution \[ u(x, y) = C \left( \frac{x}{y} \right)^{\lambda} \] is valid for arbitrary constants \( C \) and \( \lambda \). --- ### **Possible source of confusion or "what's wrong":** - The problem's structure suggests the solution is **a family of functions** depending on the ratio \( x/y \), which is consistent with the form obtained. - There may have been a typo or misstatement in the original problem, but based on the PDE provided, the separation of variables method works correctly. --- ### **Summary:** The solution for Problem 2 is **correct** and consistent with the PDE. The key point is recognizing the homogeneous form and using the substitution \( u = X(x) Y(y) \), leading to the general solution involving the ratio \( x/y \). --- If you have further questions or specific concerns about the problem, feel free to ask!