Find the general solution to the differential equation Put the problem in standard form. \( x^{2}-3 x y+x \frac{d y}{d x}=0 \) Find the integrating factor, \( \rho(x)= \) Find \( y(x)= \) Use \( C \) as the unknown constant.

We start with the differential equation \[ x^{2} - 3xy + x\frac{dy}{dx} = 0. \] **Step 1. Put the equation in standard form** Since \(x \neq 0\), we can divide the entire equation by \(x\): \[ \frac{x^{2}}{x} - 3y + \frac{dy}{dx} = 0 \quad \Longrightarrow \quad x - 3y + \frac{dy}{dx} = 0. \] Rearrange to isolate \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = 3y - x. \] Write this in standard linear form: \[ \frac{dy}{dx} - 3y = -x. \] Here, the coefficient function (in the standard linear form \(\frac{dy}{dx} + p(x)y = q(x)\)) is \[ p(x) = -3. \] --- **Step 2. Find the integrating factor, \(\rho(x)\)** The integrating factor is given by \[ \rho(x)= e^{\int p(x) \, dx} = e^{\int -3 \, dx} = e^{-3x}. \] Thus, \[ \rho(x) = e^{-3x}. \] --- **Step 3. Solve the differential equation** Multiply the standard form by the integrating factor: \[ e^{-3x}\frac{dy}{dx} - 3e^{-3x}y = -x e^{-3x}. \] Notice that the left-hand side is the derivative of \(\left(e^{-3x}y\right)\): \[ \frac{d}{dx}\left(e^{-3x}y\right) = -x e^{-3x}. \] Now integrate both sides with respect to \(x\): \[ \int \frac{d}{dx}\left(e^{-3x}y\right) \, dx = \int -x e^{-3x} \, dx. \] This gives \[ e^{-3x}y = \int -x e^{-3x} \, dx. \] **Step 3a. Evaluate the integral \(\int -x e^{-3x} \, dx\)** Let \[ I = \int -x e^{-3x} \, dx. \] Use integration by parts where \[ u = -x, \quad dv = e^{-3x}dx. \] Then \[ du = -dx, \quad v = \int e^{-3x}dx = -\frac{1}{3}e^{-3x}. \] Now, applying integration by parts: \[ I = u v - \int v\,du = \left(-x\right)\left(-\frac{1}{3}e^{-3x}\right) - \int \left(-\frac{1}{3}e^{-3x}\right)(-dx). \] Simplify the terms: \[ I = \frac{x}{3}e^{-3x} - \int \frac{1}{3}e^{-3x}\,dx. \] Now, integrate the remaining integral: \[ \int e^{-3x}\,dx = -\frac{1}{3}e^{-3x}. \] Thus, \[ I = \frac{x}{3}e^{-3x} - \frac{1}{3}\left(-\frac{1}{3}e^{-3x}\right) = \frac{x}{3}e^{-3x} + \frac{1}{9}e^{-3x}. \] So we have \[ e^{-3x}y = e^{-3x}\left(\frac{x}{3} + \frac{1}{9}\right) + C, \] where \(C\) is the constant of integration. --- **Step 4. Solve for \(y\)** Multiply both sides of the equation by \(e^{3x}\) to solve for \(y\): \[ y = \frac{x}{3} + \frac{1}{9} + C e^{3x}. \] --- **Final Answers** 1. The standard form of the differential equation is \[ \frac{dy}{dx} - 3y = -x. \] 2. The integrating factor is \[ \rho(x)= e^{-3x}. \] 3. The general solution is \[ y(x)= \frac{x}{3} + \frac{1}{9} + C e^{3x}. \]