We are given the differential equation $3x + y + x\frac{dy}{dx} = 0$ and asked to solve it.

1. Problem Description

We are given the differential equation $3x + y + x\frac{dy}{dx} = 0$ and asked to solve it.

2. Solution Steps

We can rewrite the differential equation as:

$x\frac{dy}{dx} = -3x - y$

$\frac{dy}{dx} = -\frac{3x}{x} - \frac{y}{x}$

$\frac{dy}{dx} = -3 - \frac{y}{x}$

Let $v = \frac{y}{x}$. Then $y = vx$. Differentiating with respect to $x$, we get

$\frac{dy}{dx} = v + x\frac{dv}{dx}$

Substituting into the differential equation, we have

$v + x\frac{dv}{dx} = -3 - v$

$x\frac{dv}{dx} = -3 - 2v$

$\frac{dv}{dx} = \frac{-3-2v}{x}$

$\frac{dv}{-3-2v} = \frac{dx}{x}$

$\int \frac{dv}{-3-2v} = \int \frac{dx}{x}$

$-\frac{1}{2} \int \frac{-2dv}{-3-2v} = \int \frac{dx}{x}$

$-\frac{1}{2} \ln |-3-2v| = \ln |x| + C_1$

$\ln |-3-2v| = -2\ln|x| + C_2$, where $C_2 = -2C_1$

$\ln |-3-2v| = \ln|x^{-2}| + C_2$

$|-3-2v| = e^{\ln|x^{-2}| + C_2} = e^{\ln|x^{-2}|}e^{C_2} = |x^{-2}|e^{C_2}$

$-3 - 2v = \pm e^{C_2} x^{-2} = Cx^{-2}$, where $C = \pm e^{C_2}$.

$-3 - 2(\frac{y}{x}) = \frac{C}{x^2}$

$-3 - \frac{2y}{x} = \frac{C}{x^2}$

Multiplying by $x^2$:

$-3x^2 - 2xy = C$

$3x^2 + 2xy = -C$

Let $-C = K$

$3x^2 + 2xy = K$

3. Final Answer

$3x^2 + 2xy = K$