Let's solve the boundary value problem using **separation of variables**: Given PDE and initial/boundary condition: \[ \frac{\partial u}{\partial x} = 4 \frac{\partial u}{\partial y}, \quad u(0, y) = 5e^{-2y} \] --- ### **Step 1: Assume Separable Solution** Assume \( u(x, y) = X(x)Y(y) \). --- ### **Step 2: Substitute into the PDE** \[ \frac{\partial u}{\partial x} = X'(x)Y(y) \] \[ \frac{\partial u}{\partial y} = X(x)Y'(y) \] So, \[ X'(x)Y(y) = 4 X(x)Y'(y) \] --- ### **Step 3: Separate Variables** Divide both sides by \( X(x)Y(y) \) (assuming they are nonzero): \[ \frac{X'(x)}{X(x)} = 4 \frac{Y'(y)}{Y(y)} \] Set both sides equal to a separation constant, say \( -\lambda \): \[ \frac{X'(x)}{X(x)} = 4 \frac{Y'(y)}{Y(y)} = -\lambda \] So, \[ X'(x) + \lambda X(x) = 0 \] \[ Y'(y) + \frac{\lambda}{4} Y(y) = 0 \] --- ### **Step 4: Solve the ODEs** #### For \( X(x) \): \[ X'(x) + \lambda X(x) = 0 \implies X(x) = A e^{-\lambda x} \] #### For \( Y(y) \): \[ Y'(y) + \frac{\lambda}{4} Y(y) = 0 \implies Y(y) = B e^{-\frac{\lambda}{4} y} \] --- ### **Step 5: General Solution** \[ u(x, y) = X(x)Y(y) = C e^{-\lambda x} e^{-\frac{\lambda}{4} y} \] where \( C = AB \). --- ### **Step 6: Apply Boundary Condition** Given \( u(0, y) = 5 e^{-2y} \): \[ u(0, y) = C e^{-0} e^{-\frac{\lambda}{4} y} = C e^{-\frac{\lambda}{4} y} \] Set this equal to the boundary condition: \[ C e^{-\frac{\lambda}{4} y} = 5 e^{-2y} \] This matches if \( \frac{\lambda}{4} = 2 \implies \lambda = 8 \), and \( C = 5 \). --- ### **Step 7: Write Final Solution** \[ u(x, y) = 5 e^{-8x} e^{-2y} \] Or, \[ \boxed{u(x, y) = 5 e^{-8x - 2y}} \] --- **Final Answer:** \[ \boxed{u(x, y) = 5 e^{-8x - 2y}} \]

Let's revisit the problem carefully, especially the boundary condition, to ensure accuracy. --- ### Given: - PDE: \(\frac{\partial u}{\partial x} = 4 \frac{\partial u}{\partial y}\) - Boundary condition: \(u(0, y) = 5 e^{-5 y}\) --- ### **Step 1: Assume a separable solution** Let: \[ u(x, y) = X(x) Y(y) \] --- ### **Step 2: Substitute into the PDE** \[ X'(x) Y(y) = 4 X(x) Y'(y) \] Divide both sides by \(X(x) Y(y)\): \[ \frac{X'(x)}{X(x)} = 4 \frac{Y'(y)}{Y(y)} = -\lambda \] Set both sides equal to a separation constant \(-\lambda\). --- ### **Step 3: Solve the ODEs** - For \(X(x)\): \[ X'(x) + \lambda X(x) = 0 \Rightarrow X(x) = A e^{-\lambda x} \] - For \(Y(y)\): \[ Y'(y) + \frac{\lambda}{4} Y(y) = 0 \Rightarrow Y(y) = B e^{-\frac{\lambda}{4} y} \] --- ### **Step 4: General solution** \[ u(x, y) = C e^{-\lambda x} e^{-\frac{\lambda}{4} y} \] where \(C = AB\). --- ### **Step 5: Apply the boundary condition** Given: \[ u(0, y) = 5 e^{-5 y} \] At \(x=0\): \[ u(0, y) = C e^{0} e^{-\frac{\lambda}{4} y} = C e^{-\frac{\lambda}{4} y} \] Set equal to the boundary condition: \[ C e^{-\frac{\lambda}{4} y} = 5 e^{-5 y} \] Matching exponents: \[ -\frac{\lambda}{4} y = -5 y \Rightarrow \frac{\lambda}{4} = 5 \Rightarrow \lambda = 20 \] Correspondingly, \(C = 5\). --- ### **Step 6: Final solution** \[ u(x, y) = C e^{-\lambda x} e^{-\frac{\lambda}{4} y} = 5 e^{-20 x} e^{-5 y} \] --- ### **### **Final answer:** \[ \boxed{ \boxed{ u(x, y) = 5 e^{-20 x - 5 y} } } \] --- **Summary:** - The key correction was in the boundary condition \(u(0, y) = 5 e^{-5 y}\), which led to \(\lambda = 20\), not 8. - The solution is: \[ u(x, y) = 5 e^{-20 x - 5 y} \] This satisfies both the PDE and the specified boundary condition.