Let's solve **Problem 1** step by step using the **Separation of Variables** technique. --- ### **Problem 1** \[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} \] --- ### **Step 1: Assume a Product Solution** Assume \( u(x, y) = X(x)Y(y) \). --- ### **Step 2: Substitute into the PDE** Calculate the partial derivatives: - \(\frac{\partial u}{\partial x} = X'(x)Y(y)\) - \(\frac{\partial u}{\partial y} = X(x)Y'(y)\) Substitute into the PDE: \[ X'(x)Y(y) = X(x)Y'(y) \] --- ### **Step 3: Separate Variables** Divide both sides by \( X(x)Y(y) \) (assuming neither is zero): \[ \frac{X'(x)}{X(x)} = \frac{Y'(y)}{Y(y)} \] Both sides must be equal to a constant (say \( \lambda \)), so: \[ \frac{X'(x)}{X(x)} = \lambda, \quad \frac{Y'(y)}{Y(y)} = \lambda \] --- ### **Step 4: Solve the ODEs** #### For \( X(x) \): \[ X'(x) = \lambda X(x) \] \[ \implies X(x) = A e^{\lambda x} \] #### For \( Y(y) \): \[ Y'(y) = \lambda Y(y) \] \[ \implies Y(y) = B e^{\lambda y} \] --- ### **Step 5: Write the General Product Solution** \[ u(x, y) = X(x)Y(y) = (A e^{\lambda x})(B e^{\lambda y}) = C e^{\lambda x} e^{\lambda y} \] where \( C = AB \). Or, \[ u(x, y) = C e^{\lambda(x+y)} \] --- ### **Final Answer for Problem 1:** \[ \boxed{ u(x, y) = C e^{\lambda (x + y)} } \] where \( C \) and \( \lambda \) are arbitrary constants. --- Would you like to see step-by-step solutions for the next problems? If yes, please specify which one!

Certainly! Let's proceed with solving all the given PDEs step-by-step using the method of separation of variables where applicable. --- ### **Problem 1** \[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} \] **Solution:** - **Assumption:** \( u(x,y) = X(x)Y(y) \) - **Derivatives:** \[ u_x = X'(x)Y(y), \quad u_y = X(x)Y'(y) \] - **Substitute into PDE:** \[ X'(x)Y(y) = X(x)Y'(y) \] - **Separate variables:** \[ \frac{X'(x)}{X(x)} = \frac{Y'(y)}{Y(y)} = \lambda \] - **Solve ODEs:** \[ X'(x) = \lambda X(x) \Rightarrow X(x) = A e^{\lambda x} \] \[ Y'(y) = \lambda Y(y) \Rightarrow Y(y) = B e^{\lambda y} \] - **General solution:** \[ u(x, y) = C e^{\lambda (x + y)}, \quad C = AB \] --- ### **Problem 2** \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \] **Solution:** - **Assumption:** \( u(x,y) = X(x)Y(y) \) - **Derivatives:** \[ u_x = X'(x)Y(y), \quad u_y = X(x)Y'(y) \] - **Substitute:** \[ x X'(x)Y(y) + y X(x)Y'(y) = 0 \] - **Divide through by \( XY \):** \[ x \frac{X'(x)}{X(x)} + y \frac{Y'(y)}{Y(y)} = 0 \] - **Set:** \[ x \frac{X'(x)}{X(x)} = - y \frac{Y'(y)}{Y(y)} = \mu \] - **Separate ODEs:** \[ x \frac{X'(x)}{X(x)} = \mu \Rightarrow \frac{X'(x)}{X(x)} = \frac{\mu}{x} \Rightarrow X(x) = D x^{\mu} \] \[ y \frac{Y'(y)}{Y(y)} = - \mu \Rightarrow \frac{Y'(y)}{Y(y)} = - \frac{\mu}{y} \Rightarrow Y(y) = E y^{-\mu} \] - **Solution:** \[ u(x, y) = F x^{\mu} y^{-\mu} = F \left( \frac{x}{y} \right)^{\mu} \] where \( F \) is arbitrary. --- ### **Problem 3** \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \] **This is Laplace's equation.** **Solution:** - **Assumption:** \( u(x,y) = X(x)Y(y) \) - **Substitute:** \[ X''(x)Y(y) + X(x)Y''(y) = 0 \] - **Divide through by \( XY \):** \[ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} = 0 \] - **Set:** \[ \frac{X''(x)}{X(x)} = - \lambda, \quad \frac{Y''(y)}{Y(y)} = \lambda \] - **Solve ODEs:** \[ X'' + \lambda X = 0 \Rightarrow X(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \] \[ Y'' - \lambda Y = 0 \Rightarrow Y(y) = C e^{\sqrt{\lambda} y} + D e^{-\sqrt{\lambda} y} \] - **General solution:** \[ u(x,y) = \sum_{n} [A_n \cos(n x) + B_n \sin(n x)] [C_n e^{n y} + D_n e^{-n y}] \] (For specific boundary conditions, coefficients are determined.) --- ### **Problem 4** \[ \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} \] **This is the heat equation.** **Solution:** - **Assumption:** \( u(x,t) = X(x) T(t) \) - **Substitute:** \[ X(x) T'(t) = k X''(x) T(t) \] - **Divide by \( XT \):** \[ \frac{T'(t)}{T(t)} = \frac{k X''(x)}{X(x)} = - \lambda \] - **Solve ODEs:** \[ T(t) = T_0 e^{-\lambda t} \] \[ X'' + \frac{\lambda}{k} X = 0 \Rightarrow X(x) = A \cos(\sqrt{\frac{\lambda}{k}} x) + B \sin(\sqrt{\frac{\lambda}{k}} x) \] - **General solution:** \[ u(x,t) = \sum_{n} \left[ A_n \cos(n x) + B_n \sin(n x) \right] e^{-k n^2 t} \] --- ### **Problem 5** \[ \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0 \] **Solution:** - **Assumption:** \( u(x,y) = X(x)Y(y) \) - **Derivatives:** \[ u_x = X'(x)Y(y), \quad u_y = X(x)Y'(y) \] - **Substitute:** \[ X'(x)Y(y) + X(x)Y'(y) = 0 \] - **Divide through:** \[ \frac{X'(x)}{X(x)} = - \frac{Y'(y)}{Y(y)} = \mu \] - **Solve:** \[ X'(x) = \mu X(x) \Rightarrow X(x) = A e^{\mu x} \] \[ Y'(y) = - \mu Y(y) \Rightarrow Y(y) = B e^{-\mu y} \] - **Solution:** \[ u(x,y) = C e^{\mu x - \mu y} = C e^{\mu (x - y)} \] --- ### **Summary** | Problem | PDE / Equation | General Solution Type | |---|---|---| | 1 | \(\partial u / \partial x = \partial u / \partial y\) | \( u = C e^{\lambda (x + y)} \) | | 2 | \(x u_x + y u_y=0\) | \( u = F\left(\frac{x}{y}\right) \) | | 3 | Laplace's equation \(\nabla^2 u=0\) | Fourier series / eigenfunction expansion | | 4 | Heat equation \(u_t = k u_{xx}\) | Fourier series / exponential decay | | 5 | \( u_x + u_y = 0 \) | \( u = C e^{\mu (x - y)} \) | --- Would you like further detailed solutions for specific boundary conditions or more explanations?

Certainly! Let's proceed with solving problems 6 to 10 step-by-step, using the separation of variables method where applicable. I'll include detailed explanations for each. --- ### **Problem 6** \[ x^2 \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0 \] --- ### **Step 1: Assume a product solution** \[ u(x, y) = X(x) Y(y) \] ### **Step 2: Compute derivatives** \[ u_{xx} = X''(x) Y(y), \quad u_{yy} = X(x) Y''(y) \] ### **Step 3: Substitute into PDE** \[ x^2 X''(x) Y(y) + y^2 X(x) Y''(y) = 0 \] Divide through by \( X(x) Y(y) \): \[ x^2 \frac{X''(x)}{X(x)} + y^2 \frac{Y''(y)}{Y(y)} = 0 \] --- ### **Step 4: Separate variables** Set: \[ x^2 \frac{X''(x)}{X(x)} = - y^2 \frac{Y''(y)}{Y(y)} = \lambda \] Note: Because the variables are multiplied by their squares, the separation constant depends on \(x\) and \(y\). To proceed, consider the form: \[ \frac{X''(x)}{X(x)} = \frac{\lambda}{x^{2}} \] \[ \frac{Y''(y)}{Y(y)} = - \frac{\lambda}{y^{2}} \] --- ### **Step 5: Solve the resulting ODEs** - For \(X(x)\): \[ X''(x) = \frac{\lambda}{x^{2}} X(x) \] - For \(Y(y)\): \[ Y''(y) = - \frac{\lambda}{y^{2}} Y(y) \] These are **Euler-Cauchy type** equations, or can be transformed into Bessel equations by substitution. --- ### **Step 6: Recognize the form** The general solutions involve Bessel functions of order \( \nu \), where: \[ \nu^{2} = \lambda \] The solutions are: \[ X(x) = A_{1} x^{\nu} + A_{2} x^{-\nu} \] \[ Y(y) = B_{1} y^{\nu} + B_{2} y^{-\nu} \] --- ### **### Final solution:** \[ u(x, y) = \left( A_{1} x^{\nu} + A_{2} x^{-\nu} \right) \left( B_{1} y^{\nu} + B_{2} y^{-\nu} \right) \] where \( \nu = \sqrt{\lambda} \). --- ### **Problem 7** \[ x^2 \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} + u = 0 \] --- ### **Solution:** This is similar to problem 6 with an added \( u \) term. Separation of variables proceeds similarly, assuming: \[ u(x, y) = X(x) Y(y) \] Substitute into PDE: \[ x^2 X''(x) Y(y) + y^2 X(x) Y''(y) + X(x) Y(y) = 0 \] Divide through by \( X(x) Y(y) \): \[ x^2 \frac{X''(x)}{X(x)} + y^2 \frac{Y''(y)}{Y(y)} + 1 = 0 \] Set: \[ x^2 \frac{X''(x)}{X(x)} = -\lambda \] \[ y^2 \frac{Y''(y)}{Y(y)} = -\mu \] and note that: \[ -\lambda - \mu + 1 = 0 \] \[ \Rightarrow \lambda + \mu = 1 \] The equations for \(X\) and \(Y\): \[ X''(x) = - \frac{\lambda}{x^{2}} X(x) \] \[ Y''(y) = - \frac{\mu}{y^{2}} Y(y) \] Solutions involve power functions similar to problem 6, with relations between \(\lambda\) and \(\mu\) constrained by the sum being 1. --- ### **Problem 8** \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + u = 0 \] --- ### **Solution:** - **Assumption:** \( u(x, y) = X(x) Y(y) \) - **Derivatives:** \[ u_x = X'(x)Y(y), \quad u_y = X(x)Y'(y) \] - **Substitute:** \[ x X'(x) Y(y) + y X(x) Y'(y) + X(x) Y(y) = 0 \] Divide through by \( X(x) Y(y) \): \[ x \frac{X'(x)}{X(x)} + y \frac{Y'(y)}{Y(y)} + 1 = 0 \] Set: \[ x \frac{X'(x)}{X(x)} = p, \quad y \frac{Y'(y)}{Y(y)} = q \] Then: \[ p + q + 1 = 0 \] Expressed as: \[ \frac{X'(x)}{X(x)} = \frac{p}{x} \Rightarrow X(x) = C x^{p} \] \[ \frac{Y'(y)}{Y(y)} = \frac{q}{y} \Rightarrow Y(y) = D y^{q} \] and \[ p + q + 1 = 0 \Rightarrow q = -p - 1 \] The solution: \[ u(x, y) = E x^{p} y^{-p-1} \] where \( p \) is arbitrary, so the general form is: \[ u(x,y) = \text{constant} \times x^{p} y^{-p - 1} \] --- ### **Problem 9** \[ k \frac{\partial^2 u}{\partial x^2} = p \frac{\partial u}{\partial t} \] --- ### **Solution:** This is a diffusion equation with separation of variables: \[ u(x, t) = X(x) T(t) \] Substitute: \[ k X''(x) T(t) = p X(x) T'(t) \] Divide both sides by \( X(x) T(t) \): \[ \frac{k X''(x)}{X(x)} = \frac{p T'(t)}{T(t)} = -\lambda \] Set: \[ X''(x) + \frac{\lambda}{k} X(x) = 0 \] \[ T'(t) + \frac{\lambda}{p} T(t) = 0 \] Solutions: \[ X(x) = A \cos \left( \sqrt{\frac{\lambda}{k}} x \right) + B \sin \left( \sqrt{\frac{\lambda}{k}} x \right) \] \[ T(t) = C e^{-\frac{\lambda}{p} t} \] --- ### **Problem 10** \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \lambda u = 0 \] --- ### **Solution:** This is the Helmholtz equation. Assuming: \[ u(x, y) = X(x) Y(y) \] Substitute: \[ X''(x) Y(y) + X(x) Y''(y) + \lambda X(x) Y(y) = 0 \] Divide by \( X(x) Y(y) \): \[ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} + \lambda = 0 \] Set: \[ \frac{X''(x)}{X(x)} = -\mu, \quad \frac{Y''(y)}{Y(y)} = - (\lambda - \mu) \] Solutions: \[ X(x) = A \cos(\sqrt{\mu} x) + B \sin(\sqrt{\mu} x) \] \[ Y(y) = C \cos(\sqrt{\lambda - \mu} y) + D \sin(\sqrt{\lambda - \mu} y) \] The general solution is a superposition of such eigenfunctions. --- ## **Summary of solutions for 6–10:** | Problem | Equation / Form | General Solution Type | |---|---|---| | 6 | \(x^2 u_{xx} + y^2 u_{yy} = 0\) | Power functions involving \(x^{\pm \nu} y^{\pm \nu}\) | | 7 | \(x^2 u_{xx} + y^2 u_{yy} + u=0\) | Similar to 6 with eigenvalue relations | | 8 | \(x u_x + y u_y + u=0\) | \(u = C x^{p} y^{-p-1}\) | | 9 | \(k u_{xx} = p u_t\) | Fourier modes with exponential decay in \(t\) | | 10 | \(u_{xx} + u_{yy} + \lambda u=0\) | Eigenfunction expansions with sines and cosines | --- Would you like further explanations or specific boundary/initial conditions to find particular solutions?

Certainly! Let's carefully analyze the given problem and the equations from the image, ensuring accuracy and clarity. --- ### **Understanding the Given Problem and Equations** From the image, the problem involves solving partial differential equations (PDEs) using **separation of variables** and looking for **product solutions**. The equations seem to be numbered from 6 to 10, with some specific PDEs. Here's an attempt to interpret and write down the equations correctly: --- ### **Reconstructed List of Problems** **Problem 6:** \[ x^2 \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0 \] *(Homogeneous PDE involving second derivatives)* --- **Problem 7:** \[ x^2 \frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} + u = 0 \] *(Similar to problem 6 but with an added \( u \) term)* --- **Problem 8:** \[ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + u = 0 \] *(First order PDE with variable coefficients)* --- **Problem 9:** \[ k \frac{\partial^2 u}{\partial x^2} = p \frac{\partial u}{\partial t} \] *(Heat/diffusion type equation)* --- **Problem 10:** \[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \lambda u = 0 \] *(Helmholtz equation)* --- ### **Step-by-step solution for each problem** --- ## **Problem 6:** \[ x^2 u_{xx} + y^2 u_{yy} = 0 \] --- ### **Method:** Assume **product solution**: \[ u(x, y) = X(x) Y(y) \] ### **Derivatives:** \[ u_{xx} = X''(x) Y(y), \quad u_{yy} = X(x) Y''(y) \] ### **Substitute into PDE:** \[ x^2 X''(x) Y(y) + y^2 X(x) Y''(y) = 0 \] Divide through by \( X(x) Y(y) \): \[ x^2 \frac{X''(x)}{X(x)} + y^2 \frac{Y''(y)}{Y(y)} = 0 \] ### **Separation of variables:** Set: \[ x^2 \frac{X''(x)}{X(x)} = - y^2 \frac{Y''(y)}{Y(y)} = \lambda \] Note that the separation constant \(\lambda\) may depend on \(x\) and \(y\), but for the separation to hold, both sides equal to a **constant**. --- ### **Rearranged equations:** \[ X''(x) - \frac{\lambda}{x^2} X(x) = 0 \] \[ Y''(y) + \frac{\lambda}{y^2} Y(y) = 0 \] These are **Euler-Cauchy** type equations, which have solutions in terms of power functions: \[ X(x) = A x^{m}, \quad Y(y) = B y^{n} \] where \(m, n\) satisfy: \[ m(m-1) = \lambda, \quad n(n-1) = -\lambda \] Alternatively, solutions involve Bessel functions if \(\lambda\) is specified. --- ### **Summary:** The general solution involves power functions or Bessel functions depending on boundary conditions. --- ## **Problem 7:** \[ x^2 u_{xx} + y^2 u_{yy} + u = 0 \] --- ### **Method:** Similar to problem 6, assume: \[ u = X(x) Y(y) \] Substitute derivatives: \[ x^2 X''(x) Y(y) + y^2 X(x) Y''(y) + X(x) Y(y) = 0 \] Divide through by \(X(x) Y(y)\): \[ x^2 \frac{X''(x)}{X(x)} + y^2 \frac{Y''(y)}{Y(y)} + 1 = 0 \] Set: \[ x^2 \frac{X''(x)}{X(x)} = -\lambda \] \[ y^2 \frac{Y''(y)}{Y(y)} = -\mu \] with the condition: \[ -\lambda - \mu + 1 = 0 \quad \Rightarrow \quad \lambda + \mu = 1 \] The equations become: \[ X''(x) + \frac{\lambda}{x^{2}} X(x) = 0 \] \[ Y''(y) + \frac{\mu}{y^{2}} Y(y) = 0 \] Solutions are similar to problem 6 involving power functions or Bessel functions, with the constraint \(\lambda + \mu = 1\). --- ## **Problem 8:** \[ x u_x + y u_y + u = 0 \] --- ### **Method:** Assume: \[ u(x, y) = X(x) Y(y) \] Compute derivatives: \[ u_x = X'(x) Y(y), \quad u_y = X(x) Y'(y) \] Substitute into PDE: \[ x X'(x) Y(y) + y X(x) Y'(y) + X(x) Y(y) = 0 \] Divide through by \(X(x) Y(y)\): \[ x \frac{X'(x)}{X(x)} + y \frac{Y'(y)}{Y(y)} + 1 = 0 \] Set: \[ x \frac{X'(x)}{X(x)} = p, \quad y \frac{Y'(y)}{Y(y)} = q \] Thus: \[ p + q + 1 = 0 \quad \Rightarrow \quad q = -p - 1 \] Solve: \[ X'(x) = \frac{p}{x} X(x) \Rightarrow X(x) = A x^{p} \] \[ Y'(y) = \frac{q}{y} Y(y) \Rightarrow Y(y) = B y^{q} = B y^{-p - 1} \] **Solution:** \[ u(x, y) = C x^{p} y^{-p - 1} \] where \(p\) is an arbitrary parameter. --- ## **Problem 9:** \[ k u_{xx} = p u_t \] --- ### **Method:** Separable solution: \[ u(x, t) = X(x) T(t) \] Substitute: \[ k X''(x) T(t) = p X(x) T'(t) \] Divide by \(X(x) T(t)\): \[ \frac{k X''(x)}{X(x)} = \frac{p T'(t)}{T(t)} = -\lambda \] Thus: \[ X''(x) + \frac{\lambda}{k} X(x) = 0 \] \[ T'(t) + \frac{\lambda}{p} T(t) = 0 \] Solutions: \[ X(x) = A \cos \left( \sqrt{\frac{\lambda}{k}} x \right) + B \sin \left( \sqrt{\frac{\lambda}{k}} x \right) \] \[ T(t) = C e^{-\frac{\lambda}{p} t} \] --- ## **Problem 10:** \[ u_{xx} + u_{yy} + \lambda u = 0 \] --- ### **Method:** Assume: \[ u(x, y) = X(x) Y(y) \] Substitute: \[ X''(x) Y(y) + X(x) Y''(y) + \lambda X(x) Y(y) = 0 \] Divide by \(X Y\): \[ \frac{X''(x)}{X(x)} + \frac{Y''(y)}{Y(y)} + \lambda = 0 \] Set: \[ \frac{X''(x)}{X(x)} = - \mu \] \[ \frac{Y''(y)}{Y(y)} = - (\lambda - \mu) \] Solutions are: \[ X(x) = A \cos (\sqrt{\mu} x) + B \sin (\sqrt{\mu} x) \] \[ Y(y) = C \cos (\sqrt{\lambda - \mu} y) + D \sin (\sqrt{\lambda - \mu} y) \] --- ## **Summary:** - The **equations** involve eigenfunction expansions, power functions, or Bessel functions depending on boundary conditions. - The **solutions** depend on the specific form of the PDEs and boundary conditions. --- ### **Conclusion:** The original question was to **use separation of variables** to find **product solutions** for the specified PDEs. The above steps provide accurate and detailed solutions based on the reconstructed equations. --- If you have **specific boundary or initial conditions**, I can help you find **particular solutions** tailored to those conditions.