Let's solve each question step-by-step: --- ### **1. Separation of Variables Boundary Value Problem** Given: \[ \frac{\partial u}{\partial x} = 4 \frac{\partial u}{\partial y}, \qquad u(0, y) = 5 e^{-2y} \] **Step 1:** Assume a solution of the form \( u(x, y) = X(x) Y(y) \). **Step 2:** Substitute into the PDE: \[ \frac{\partial u}{\partial x} = X'(x) Y(y), \qquad \frac{\partial u}{\partial y} = X(x) Y'(y) \] So, \[ X'(x) Y(y) = 4 X(x) Y'(y) \] \[ \frac{X'(x)}{4 X(x)} = \frac{Y'(y)}{Y(y)} = \lambda \quad (\text{separation constant}) \] **Step 3:** Solve each ODE: - \( X'(x) = 4\lambda X(x) \implies X(x) = A e^{4\lambda x} \) - \( Y'(y) = \lambda Y(y) \implies Y(y) = B e^{\lambda y} \) General solution: \[ u(x, y) = C e^{4\lambda x} e^{\lambda y} \] **Step 4:** Use the boundary condition \( u(0, y) = 5 e^{-2y} \): \[ u(0, y) = C e^{0} e^{\lambda y} = C e^{\lambda y} = 5 e^{-2y} \] So, \( \lambda = -2 \), \( C = 5 \). **Final Answer:** \[ \boxed{u(x, y) = 5 e^{4(-2)x} e^{-2y} = 5 e^{-8x - 2y}} \] --- ### **2. Homogeneous Differential Equation (Substitution Method)** Given: \[ yz\,dx - zx\,dy - y^2\,dz = 0 \] **Step 1:** Write in terms of differentials: \[ yz\,dx - zx\,dy - y^2\,dz = 0 \] **Step 2:** Express as a ratio: \[ \frac{dx}{x} - \frac{dy}{y} - \frac{dz}{z} = 0 \] **Step 3:** Rearranging: \[ \frac{dx}{x} - \frac{dy}{y} = \frac{dz}{z} \] Integrate both sides: \[ \int \left( \frac{dx}{x} - \frac{dy}{y} \right) = \int \frac{dz}{z} \] \[ \ln|x| - \ln|y| = \ln|z| + C \] Combine the logs: \[ \ln\left|\frac{x}{y}\right| = \ln|z| + C \] Exponentiate both sides: \[ \frac{x}{y} = Kz \quad (K = e^C) \] **Final Answer:** \[ \boxed{\frac{x}{y} = Kz} \] or \[ \boxed{x = K y z} \] where \( K \) is an arbitrary constant. --- ### **3. Laplace Transform of \( \frac{\partial^2 u(x, t)}{\partial t^2} \)** Let \( \mathcal{L}[u(x, t)] = U(x, s) \). Laplace Transform property: \[ \mathcal{L}\left[\frac{\partial^2 u(x, t)}{\partial t^2}\right] = s^2 U(x, s) - s u(x, 0) - u_t(x, 0) \] **Final Answer:** \[ \boxed{\mathcal{L}\left[\frac{\partial^2 u(x, t)}{\partial t^2}\right] = s^2 U(x, s) - s u(x, 0) - u_t(x, 0)} \] --- If you need further clarifications or explanations, let me know!

Certainly! Let's go through each part step-by-step with detailed calculations and explanations. --- ## **Part 1: Solving the Boundary Value Problem via Separation of Variables** ### **Given:** \[ \frac{\partial u}{\partial x} = 4 \frac{\partial u}{\partial y} \] with boundary condition: \[ u(0, y) = 5 e^{-2y} \] --- ### **Step 1: Assume a separable solution** Suppose: \[ u(x, y) = X(x) Y(y) \] ### **Step 2: Substitute into the PDE** Calculate derivatives: \[ \frac{\partial u}{\partial x} = X'(x) Y(y) \] \[ \frac{\partial u}{\partial y} = X(x) Y'(y) \] Plug into PDE: \[ X'(x) Y(y) = 4 X(x) Y'(y) \] Divide both sides by \( X(x) Y(y) \) (assuming these functions are non-zero): \[ \frac{X'(x)}{X(x)} = 4 \frac{Y'(y)}{Y(y)} = \lambda \quad \text{(separation constant)} \] --- ### **Step 3: Solve the ODEs** - For \( X(x) \): \[ X'(x) = \lambda X(x) \implies \frac{dX}{X} = \lambda dx \] \[ X(x) = A e^{\lambda x} \] - For \( Y(y) \): \[ Y'(y) = \frac{\lambda}{4} Y(y) \implies \frac{dY}{Y} = \frac{\lambda}{4} dy \] \[ Y(y) = B e^{\frac{\lambda}{4} y} \] --- ### **Step 4: General solution** \[ u(x,y) = X(x) Y(y) = C e^{\lambda x} e^{\frac{\lambda}{4} y} \] where \( C = AB \). --- ### **Step 5: Apply boundary condition** At \( x=0 \): \[ u(0, y) = C e^{0} e^{\frac{\lambda}{4} y} = C e^{\frac{\lambda}{4} y} = 5 e^{-2 y} \] Comparing: \[ C e^{\frac{\lambda}{4} y} = 5 e^{-2 y} \] For this to hold for all \( y \): \[ \frac{\lambda}{4} y = -2 y \implies \frac{\lambda}{4} = -2 \implies \lambda = -8 \] And: \[ C = 5 \] ### **Final solution:** \[ u(x, y) = 5 e^{-8 x} e^{\frac{-8}{4} y} = 5 e^{-8 x} e^{-2 y} \] **Answer:** \[ \boxed{ u(x, y) = 5 e^{-8 x - 2 y} } \] --- ## **Part 2: Homogeneous Differential Equation** Given: \[ y z\, dx - z x\, dy - y^2\, dz = 0 \] --- ### **Step 1: Rewrite the differential equation** Divide through by \( xyz \) (assuming variables are non-zero): \[ \frac{y z}{x y z} dx - \frac{z x}{x y z} dy - \frac{y^2}{x y z} dz = 0 \] which simplifies to: \[ \frac{1}{x} dx - \frac{1}{y} dy - \frac{y}{x z} dz = 0 \] However, it's better to consider the form in differentials directly: \[ y z\, dx - z x\, dy - y^2\, dz = 0 \] --- ### **Step 2: Express as a ratio** Rearranged: \[ y z\, dx = z x\, dy + y^2\, dz \] Divide through by \( y z \): \[ dx = \frac{z x}{y z} dy + \frac{y^2}{y z} dz \] \[ dx = \frac{x}{y} dy + \frac{y}{z} dz \] Now, write as: \[ dx - \frac{x}{y} dy - \frac{y}{z} dz = 0 \] --- ### **Step 3: Recognize the form** This suggests the substitution: \[ \frac{dx}{x} - \frac{dy}{y} = \frac{dz}{z} \] **Alternatively**, it's more straightforward to notice that the original differential equation is homogeneous of degree 2 (since each term is of degree 2 in variables). Let's try substitution: \[ v = \frac{y}{x} \] which implies: \[ y = v x \] and \[ dy = v dx + x dv \] --- ### **Step 4: Substitute into the original** Recall: \[ y z\, dx - z x\, dy - y^2\, dz = 0 \] Express \( y \) and \( dy \): \[ (v x) z\, dx - z x (v dx + x dv) - (v x)^2 dz = 0 \] Expand: \[ v x z\, dx - v z x\, dx - z x^2 dv - v^2 x^2 dz = 0 \] Observe that: \[ v x z\, dx - v z x\, dx = 0 \] These cancel out, leaving: \[ - z x^2 dv - v^2 x^2 dz = 0 \] Divide through by \( x^2 \): \[ - z dv - v^2 dz = 0 \] Rearranged: \[ z dv + v^2 dz = 0 \] --- ### **Step 5: Separate variables** Rewrite: \[ z dv = - v^2 dz \] \[ \frac{dv}{v^2} = - \frac{dz}{z} \] Integrate both sides: \[ \int v^{-2} dv = - \int z^{-1} dz \] Calculate integrals: \[ \int v^{-2} dv = - v^{-1} + C_1 \] \[ - \int z^{-1} dz = - \ln |z| + C_2 \] Set: \[ - v^{-1} = - \ln |z| + C \] or \[ v^{-1} = \ln |z| + C' \] Recall \( v = y/x \), so: \[ \frac{x}{y} = v \implies v^{-1} = \frac{y}{x} \] Thus: \[ \frac{y}{x} = \frac{1}{\ln |z| + C'} \] or equivalently, \[ \boxed{ \frac{y}{x} = \frac{1}{\ln |z| + C} } \] which describes the general solution. Alternatively, write: \[ x = y \left( \ln |z| + C \right) \] --- ## **Part 3: Laplace Transform of the second derivative** Given: \[ \mathcal{L}\left[\frac{\partial^2 u(x, t)}{\partial t^2}\right] \] ### **Recall the Laplace transform of derivatives:** \[ \mathcal{L}\left[\frac{\partial u}{\partial t}\right] = s U(x, s) - u(x, 0) \] Therefore: \[ \mathcal{L}\left[\frac{\partial^2 u}{\partial t^2}\right] = s \mathcal{L}\left[\frac{\partial u}{\partial t}\right] - \left[\frac{\partial u}{\partial t}\right]_{t=0} \] But more directly, the property: \[ \mathcal{L}\left[\frac{\partial^2 u}{\partial t^2}\right] = s^2 U(x, s) - s u(x, 0) - u_t(x, 0) \] --- ## **Summary of Final Answers:** 1. **Solution to PDE:** \[ \boxed{ u(x, y) = 5 e^{-8 x - 2 y} } \] 2. **General solution of the differential equation:** \[ \boxed{ \frac{y}{x} = \frac{1}{\ln |z| + C} } \] 3. **Laplace transform of second derivative:** \[ \boxed{ \mathcal{L}\left[\frac{\partial^2 u(x, t)}{\partial t^2}\right] = s^2 U(x, s) - s u(x, 0) - u_t(x, 0) } \] --- If you need further clarification or details, feel free to ask!