Let's solve each PDE **using separation of variables** and find product solutions where possible. ### **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 equation** \[ \frac{\partial u}{\partial x} = Y(y) X'(x) \] \[ \frac{\partial u}{\partial y} = X(x) Y'(y) \] Given, \[ Y(y) X'(x) = X(x) Y'(y) \] #### **Step 3: Separate variables** \[ \frac{X'(x)}{X(x)} = \frac{Y'(y)}{Y(y)} \] Set both sides equal to a constant, say \(\lambda\): \[ \frac{X'(x)}{X(x)} = \lambda \implies X'(x) = \lambda X(x) \] \[ \frac{Y'(y)}{Y(y)} = \lambda \implies Y'(y) = \lambda Y(y) \] #### **Step 4: Solve the ODEs** \[ X(x) = A e^{\lambda x} \] \[ Y(y) = B e^{\lambda y} \] #### **Step 5: Write the general solution** \[ u(x, y) = X(x) Y(y) = C e^{\lambda (x+y)} \] where \(C = AB\), and \(\lambda\) is any constant. **Final Answer:** \[ \boxed{u(x, y) = C e^{\lambda(x + y)}} \] --- Would you like to proceed to the next problem (#2) or do you want all four solved here?

Certainly! Let's proceed to solve the second PDE step-by-step. --- ### **2.** \(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0\) #### **Step 1: Assume a product solution** Let \(u(x, y) = X(x)Y(y)\). #### **Step 2: Substitute into the PDE** \[ \frac{\partial u}{\partial x} = X'(x) Y(y) \] \[ \frac{\partial u}{\partial y} = X(x) Y'(y) \] The PDE becomes: \[ X'(x) Y(y) + X(x) Y'(y) = 0 \] #### **Step 3: Divide through by \(X(x) Y(y)\) (assuming they are non-zero)** \[ \frac{X'(x)}{X(x)} + \frac{Y'(y)}{Y(y)} = 0 \] Set: \[ \frac{X'(x)}{X(x)} = - \frac{Y'(y)}{Y(y)} = \lambda \] where \(\lambda\) is a separation constant. --- ### **Step 4: Solve the ODEs** - For \(X(x)\): \[ X'(x) = \lambda X(x) \Rightarrow X(x) = A e^{\lambda x} \] - For \(Y(y)\): \[ Y'(y) = -\lambda Y(y) \Rightarrow Y(y) = B e^{-\lambda y} \] --- ### **Step 5: Write the general solution** \[ u(x, y) = X(x) Y(y) = A B e^{\lambda x} e^{-\lambda y} = C e^{\lambda (x - y)} \] where \(C = AB\) is an arbitrary constant, and \(\lambda\) is any real number. --- ### **Final Answer:** \[ \boxed{ u(x, y) = C e^{\lambda (x - y)} } \] --- Would you like to continue with the next PDE?

Thank you for the clarification. The correct PDE for question 2 is: \[ \frac{\partial u}{\partial x} + 3 \frac{\partial u}{\partial y} = 0 \] Let's solve this step-by-step using separation of variables and product solutions. --- ### **2.** \(\frac{\partial u}{\partial x} + 3 \frac{\partial u}{\partial y} = 0\) --- #### **Step 1: Assume a product solution** Let: \[ u(x, y) = X(x) Y(y) \] --- #### **Step 2: Compute partial derivatives** \[ \frac{\partial u}{\partial x} = X'(x) Y(y) \] \[ \frac{\partial u}{\partial y} = X(x) Y'(y) \] Substitute into the PDE: \[ X'(x) Y(y) + 3 X(x) Y'(y) = 0 \] --- #### **Step 3: Divide through by \(X(x) Y(y)\) (assuming non-zero functions)** \[ \frac{X'(x)}{X(x)} + 3 \frac{Y'(y)}{Y(y)} = 0 \] Set both sides equal to a separation constant \(\lambda\): \[ \frac{X'(x)}{X(x)} = -3 \frac{Y'(y)}{Y(y)} = \lambda \] --- #### **Step 4: Solve the ODEs** - For \(X(x)\): \[ X'(x) = \lambda X(x) \Rightarrow X(x) = A e^{\lambda x} \] - For \(Y(y)\): \[ Y'(y) = -\frac{\lambda}{3} Y(y) \Rightarrow Y(y) = B e^{-\frac{\lambda}{3} y} \] --- #### **Step 5: Write the general product solution** \[ u(x, y) = X(x) Y(y) = A B e^{\lambda x} e^{-\frac{\lambda}{3} y} = C e^{\lambda \left(x - \frac{y}{3}\right)} \] where \(C = AB\) and \(\lambda\) is an arbitrary real constant. --- ### **Final answer:** \[ \boxed{ u(x, y) = C e^{\lambda \left( x - \frac{y}{3} \right)} } \] --- Would you like to proceed to the third or fourth problem?

Certainly! Let's proceed to solve the third PDE step-by-step. --- ### **3.** \(\frac{\partial u}{\partial x} = y \frac{\partial u}{\partial y}\) --- #### **Step 1: Assume a product solution** Let: \[ u(x, y) = X(x) Y(y) \] --- #### **Step 2: Compute partial derivatives** \[ \frac{\partial u}{\partial x} = X'(x) Y(y) \] \[ \frac{\partial u}{\partial y} = X(x) Y'(y) \] Substitute into the PDE: \[ X'(x) Y(y) = y \cdot X(x) Y'(y) \] --- #### **Step 3: Divide both sides by \(X(x) Y(y)\) (assuming non-zero functions)** \[ \frac{X'(x)}{X(x)} = y \frac{Y'(y)}{Y(y)} \] Note that the left side depends only on \(x\), and the right side depends only on \(y\). For equality to hold for all \(x, y\), both sides must equal a constant, say \(\lambda\). Set: \[ \frac{X'(x)}{X(x)} = \lambda \] and \[ y \frac{Y'(y)}{Y(y)} = \lambda \] --- ### **Step 4: Solve the ODEs** - For \(X(x)\): \[ X'(x) = \lambda X(x) \Rightarrow X(x) = A e^{\lambda x} \] - For \(Y(y)\): \[ y \frac{Y'(y)}{Y(y)} = \lambda \Rightarrow \frac{Y'(y)}{Y(y)} = \frac{\lambda}{y} \] This is a separable ODE: \[ \frac{Y'(y)}{Y(y)} = \frac{\lambda}{y} \] Integrate both sides: \[ \int \frac{Y'(y)}{Y(y)} dy = \int \frac{\lambda}{y} dy \] \[ \ln |Y(y)| = \lambda \ln |y| + C \] Exponentiate: \[ Y(y) = D y^{\lambda} \] where \(D = e^{C}\) is an arbitrary constant. --- ### **Step 5: Write the general solution** \[ u(x, y) = X(x) Y(y) = A e^{\lambda x} \cdot D y^{\lambda} = C e^{\lambda x} y^{\lambda} \] where \(C = AD\). ### **Final answer:** \[ \boxed{ u(x, y) = C e^{\lambda x} y^{\lambda} } \] with \(\lambda\) an arbitrary constant. --- Would you like to proceed to the fourth PDE?

Thank you for the clarification. The correct PDE for question 4 is: \[ u_x + u_y = u \] Let's solve this PDE step-by-step using separation of variables and product solutions. --- ### **4.** \(u_x + u_y = u\) --- #### **Step 1: Assume a product solution** Let: \[ u(x, y) = X(x) Y(y) \] --- #### **Step 2: Compute partial derivatives** \[ u_x = X'(x) Y(y) \] \[ u_y = X(x) Y'(y) \] Substitute into the PDE: \[ X'(x) Y(y) + X(x) Y'(y) = X(x) Y(y) \] --- #### **Step 3: Divide through by \(X(x) Y(y)\) (assuming non-zero functions)** \[ \frac{X'(x)}{X(x)} + \frac{Y'(y)}{Y(y)} = 1 \] Note that the left side is a sum of functions of \(x\) and \(y\). To separate variables, set: \[ \frac{X'(x)}{X(x)} = \alpha \] \[ \frac{Y'(y)}{Y(y)} = 1 - \alpha \] where \(\alpha\) is a separation constant. --- ### **Step 4: Solve the ODEs** - For \(X(x)\): \[ X'(x) = \alpha X(x) \Rightarrow X(x) = A e^{\alpha x} \] - For \(Y(y)\): \[ Y'(y) = (1 - \alpha) Y(y) \Rightarrow Y(y) = B e^{(1 - \alpha) y} \] --- ### **Step 5: Write the general solution** \[ u(x, y) = X(x) Y(y) = A B e^{\alpha x} e^{(1 - \alpha) y} = C e^{\alpha x + (1 - \alpha) y} \] where \(C = AB\), and \(\alpha\) is an arbitrary constant. Alternatively, since \(\alpha\) is arbitrary, we can write the general solution as a superposition (integral over all \(\alpha\)): \[ u(x, y) = \int_{-\infty}^\infty c(\alpha) e^{\alpha x + (1 - \alpha) y} d\alpha \] But often, the solution in the separated form: \[ \boxed{ u(x, y) = C e^{\alpha x + (1 - \alpha) y} } \] with arbitrary constant \(\alpha\), suffices. --- ### **Final answer:** \[ \boxed{ u(x, y) = C e^{\alpha x + (1 - \alpha) y} } \] where \(\alpha\) is an arbitrary constant. --- **Summary of all solutions:** 1. \(u = C e^{\lambda (x + y)}\) 2. \(u = C e^{\lambda (x - y/3)}\) 3. \(u = C e^{\lambda x} y^{\lambda}\) 4. \(u = C e^{\alpha x + (1 - \alpha) y}\) Would you like further explanations or help with boundary conditions?

Thank you for the clarification. The correct PDE for question 4 is: \[ u_x + u = u_y \] Let's proceed to solve this PDE step-by-step using separation of variables and product solutions. --- ### **4.** \(u_x + u = u_y\) --- #### **Step 1: Rewrite the PDE** Rewrite the PDE to bring all derivatives to one side: \[ u_x - u_y = -u \] Alternatively, keep it as: \[ u_x + u = u_y \] --- #### **Step 2: Assume a product solution** Let: \[ u(x, y) = X(x) Y(y) \] --- #### **Step 3: Compute partial derivatives** \[ u_x = X'(x) Y(y) \] \[ u_y = X(x) Y'(y) \] Substitute into the PDE: \[ X'(x) Y(y) + X(x) Y(y) = X(x) Y'(y) \] --- #### **Step 4: Divide through by \(X(x) Y(y)\) (assuming non-zero functions)** \[ \frac{X'(x)}{X(x)} + 1 = \frac{Y'(y)}{Y(y)} \] The left side depends only on \(x\), and the right side only on \(y\). For equality to hold for all \(x, y\), both sides must be equal to a constant, say \(\lambda\): \[ \frac{X'(x)}{X(x)} + 1 = \lambda \] \[ \frac{Y'(y)}{Y(y)} = \lambda \] --- ### **Step 5: Solve the ODEs** - For \(X(x)\): \[ X'(x) = (\lambda - 1) X(x) \] \[ X(x) = A e^{(\lambda - 1) x} \] - For \(Y(y)\): \[ Y'(y) = \lambda Y(y) \Rightarrow Y(y) = B e^{\lambda y} \] --- ### **Step 6: Write the general solution** \[ u(x, y) = X(x) Y(y) = A e^{(\lambda - 1) x} \cdot B e^{\lambda y} = C e^{(\lambda - 1) x + \lambda y} \] where \(C = AB\) is an arbitrary constant, and \(\lambda\) is an arbitrary parameter. --- ### **Final answer:** \[ \boxed{ u(x, y) = C e^{(\lambda - 1) x + \lambda y} } \] --- ### **Summary:** The general solution to the PDE \(u_x + u = u_y\) is: \[ u(x, y) = C e^{(\lambda - 1) x + \lambda y} \] where \(\lambda\) is an arbitrary constant, and \(C\) is an arbitrary constant. --- Would you like further clarification or assistance with boundary conditions?

Certainly! Let's carefully solve the PDE: \[ u_x = u_y + u \] using separation of variables and product solutions. --- ### Step 1: Rewrite the PDE The PDE is: \[ u_x = u_y + u \] which can be rearranged as: \[ u_x - u_y = u \] --- ### Step 2: Assume a product solution Let: \[ u(x, y) = X(x) Y(y) \] --- ### Step 3: Compute derivatives \[ u_x = X'(x) Y(y) \] \[ u_y = X(x) Y'(y) \] Substitute into the PDE: \[ X'(x) Y(y) = X(x) Y'(y) + X(x) Y(y) \] or \[ X'(x) Y(y) = X(x) Y'(y) + X(x) Y(y) \] --- ### Step 4: Divide both sides by \(X(x) Y(y)\) \[ \frac{X'(x)}{X(x)} = \frac{Y'(y)}{Y(y)} + 1 \] Note that the left depends only on \(x\), and the right depends only on \(y\). Therefore, both sides equal a constant, say \(\lambda\): \[ \frac{X'(x)}{X(x)} = \lambda \] \[ \frac{Y'(y)}{Y(y)} + 1 = \lambda \] which implies: \[ \frac{Y'(y)}{Y(y)} = \lambda - 1 \] --- ### Step 5: Solve the ODEs - For \(X(x)\): \[ X'(x) = \lambda X(x) \Rightarrow X(x) = A e^{\lambda x} \] - For \(Y(y)\): \[ Y'(y) = (\lambda - 1) Y(y) \Rightarrow Y(y) = B e^{(\lambda - 1) y} \] --- ### **Step 6: Write the general solution** \[ u(x, y) = X(x) Y(y) = A e^{\lambda x} \cdot B e^{(\lambda - 1) y} = C e^{\lambda x + (\lambda - 1) y} \] where \(C = AB\). Alternatively, since \(\lambda\) is arbitrary, the general solution is a superposition of such solutions: \[ u(x, y) = \int c(\lambda) e^{\lambda x + (\lambda - 1) y} d\lambda \] or, in the simplified form, the family of solutions: \[ \boxed{ u(x, y) = C e^{\lambda x + (\lambda - 1) y} } \] where \(\lambda\) is an arbitrary parameter. --- ### **Summary:** The product solutions for the PDE \(u_x = u_y + u\) are of the form: \[ \boxed{ u(x, y) = C e^{\lambda x + (\lambda - 1) y} } \] with \(\lambda\) arbitrary. --- If you'd like, I can help with specific boundary conditions or further interpretations!