Let's solve each PDE step-by-step: --- ## (a) \(\frac{\partial u}{\partial x} + 3 \frac{\partial u}{\partial y} = 5u + \tan(y - 3x)\) This is a **linear first-order PDE** and can be solved by the method of characteristics. ### **Step 1: Write the characteristic equations** For PDE: \[ \frac{\partial u}{\partial x} + 3\frac{\partial u}{\partial y} = 5u + \tan(y - 3x) \] The characteristic equations are: \[ \frac{dx}{1} = \frac{dy}{3} = \frac{du}{5u + \tan(y-3x)} \] ### **Step 2: Solve the first two equations** \[ \frac{dx}{1} = \frac{dy}{3} \implies dy = 3dx \implies y - 3x = C_1 \] So, along a characteristic curve, \(y - 3x\) is constant. ### **Step 3: Solve for \(u\) along the characteristic** Let \(s\) be the parameter along the characteristic, with \(x = s\), \(y = 3s + C_1\): \[ \frac{du}{ds} = 5u + \tan(C_1) \] This is a first-order ODE in \(u\): \[ \frac{du}{ds} - 5u = \tan(C_1) \] This is a linear ODE. Integrating factor (IF) is \(e^{-5s}\): \[ e^{-5s}\frac{du}{ds} - 5e^{-5s}u = \tan(C_1)e^{-5s} \] \[ \frac{d}{ds}\left(u e^{-5s}\right) = \tan(C_1) e^{-5s} \] Integrate both sides: \[ u e^{-5s} = \int \tan(C_1) e^{-5s} ds + C_2 \] \[ u e^{-5s} = \tan(C_1) \int e^{-5s} ds + C_2 \] \[ u e^{-5s} = \tan(C_1) \left( -\frac{1}{5} e^{-5s} \right) + C_2 \] \[ u e^{-5s} = -\frac{1}{5}\tan(C_1) e^{-5s} + C_2 \] \[ u = -\frac{1}{5}\tan(C_1) + C_2 e^{5s} \] ### **Step 4: Substitute back for \(s, C_1\)** Recall \(C_1 = y - 3x\), \(s = x\): \[ u(x, y) = -\frac{1}{5} \tan(y - 3x) + f(y - 3x) e^{5x} \] where \(f\) is an arbitrary function, absorbing the constant \(C_2\). --- ## **Final Answer for (a):** \[ \boxed{ u(x, y) = -\frac{1}{5} \tan(y - 3x) + f(y - 3x) e^{5x} } \] where \(f\) is an arbitrary function. --- Would you like the step-by-step for (b) and (c) as well?

Let's proceed with solving the remaining two PDEs step-by-step. --- ## (b) \(\frac{\partial u}{\partial y} - \frac{\partial u}{\partial z} = (z + y)u + 22\) This is a **first-order linear PDE** involving variables \(y, z\). ### **Step 1: Write the PDE** \[ u_y - u_z = (z + y)u + 22 \] ### **Step 2: Find the characteristic equations** The associated characteristic equations are derived from the coefficients of derivatives: \[ \frac{dy}{1} = \frac{dz}{-1} = \frac{du}{(z + y) u + 22} \] ### **Step 3: Solve the first two equations** \[ \frac{dy}{1} = \frac{dz}{-1} \implies dy = -dz \implies y + z = C_1 \] So, along a characteristic, \(y + z = \text{constant}\). Let \(w = y + z\). Then \(z = w - y\). ### **Step 4: Rewrite the PDE in terms of \(w\)** Express \(u\) as a function of \(y\) and \(w\), i.e., \(u = u(y, w)\). Since \(w = y + z\), the derivatives: \[ u_y = \frac{\partial u}{\partial y} + \frac{\partial u}{\partial w} \frac{\partial w}{\partial y} = u_y + u_w \cdot 1 \] \[ u_z = u_w \frac{\partial w}{\partial z} = u_w \cdot 1 \] But from the original PDE, \(u_y - u_z = (z + y) u + 22\): \[ (u_y + u_w) - u_w = (w) u + 22 \] \[ u_y = w u + 22 \] Now, since \(w = y + z\), and \(u = u(y,w)\), the PDE reduces to: \[ u_y = w u + 22 \] This is a first-order ODE in \(y\) with parameter \(w\). ### **Step 5: Solve the ODE** Treat \(w\) as a parameter: \[ \frac{\partial u}{\partial y} = w u + 22 \] This is linear in \(u\): \[ u_y - w u = 22 \] The integrating factor: \[ \mu(y) = e^{-w y} \] Multiply both sides: \[ e^{-w y} u_y - w e^{-w y} u = 22 e^{-w y} \] \[ \frac{\partial}{\partial y} \left( u e^{-w y} \right) = 22 e^{-w y} \] Integrate both sides: \[ u e^{-w y} = \int 22 e^{-w y} dy + C(w) \] \[ u e^{-w y} = 22 \left( -\frac{1}{w} e^{-w y} \right) + C(w) \] \[ u e^{-w y} = -\frac{22}{w} e^{-w y} + C(w) \] Multiply through by \(e^{w y}\): \[ u = -\frac{22}{w} + C(w) e^{w y} \] Recall that \(w = y + z\), so: \[ u(y, z) = -\frac{22}{y + z} + \phi(y + z) e^{(y + z) y} \] where \(\phi(w) = C(w)\) is an arbitrary function. --- ## **Final answer for (b):** \[ \boxed{ u(y,z) = -\frac{22}{y + z} + \phi(y + z) e^{(y + z) y} } \] where \(\phi\) is an arbitrary function of \(w = y + z\). --- ## (c) \(z y - 2 u_x + y(z - 2) = 2(z - y)\) First, rewrite the PDE clearly: \[ z y - 2 u_x + y(z - 2) = 2(z - y) \] ### **Step 1: Isolate \(u_x\)** \[ -2 u_x = 2(z - y) - z y - y(z - 2) \] Simplify RHS: \[ 2(z - y) - z y - y(z - 2) \] Compute \(y(z - 2)\): \[ z y - 2 y \] Now, substitute: \[ -2 u_x = 2z - 2 y - z y - z y + 2 y \] Note that \(- y(z - 2) = - z y + 2 y\). Now, combine: \[ -2 u_x = 2z - 2 y - z y - z y + 2 y \] The \(-2 y + 2 y\) cancel out: \[ -2 u_x = 2z - 2 z y \] Divide both sides by \(-2\): \[ u_x = -z + z y \] --- ### **Step 2: Integrate with respect to \(x\)** Since RHS depends on \(z\) and \(y\), which are independent variables, for \(u_x\), treat \(z, y\) as parameters. \[ u(x, y, z) = \int (-z + z y) dx + \psi(y, z) \] \[ u(x, y, z) = (-z + z y) x + \psi(y, z) \] where \(\psi(y, z)\) is an arbitrary function of \(y, z\). --- ## **Final answer for (c):** \[ \boxed{ u(x, y, z) = x z (y - 1) + \psi(y, z) } \] where \(\psi\) is arbitrary. --- ## **Summary of all solutions:** 1. **(a)** \[ u(x, y) = -\frac{1}{5} \tan(y - 3x) + f(y - 3x) e^{5x} \] 2. **(b)** \[ u(y, z) = -\frac{22}{y + z} + \phi(y + z) e^{(y + z) y} \] 3. **(c)** \[ u(x, y, z) = x z (y - 1) + \psi(y, z) \] where \(f, \phi, \psi\) are arbitrary functions of their indicated arguments. --- If you'd like further clarification on any step, feel free to ask!