For $\frac{dy}{dx} + \frac{y}{x} = 3$, with $x > 0$, which of the following statements are true?

I. Its integrating factor is $x$.

II. Its general solution is $y = 3x + \frac{C}{x}$.

III. $y = 3x$ is one of its solutions.

IV. Its solution through $(1, 2)$ is $y = \frac{3x}{2} + \frac{1}{2x}$.

Explanation

1. 01

   I: $e^{\int \frac{dx}{x}} = x$, and $\frac{d}{dx}(xy) = 3x$ gives $xy = \frac{3x^{2}}{2} + C$, so $y = \frac{3x}{2} + \frac{C}{x}$; II has lost the $\frac{1}{2}$.

2. 02

   III: on $y = 3x$ the left side is $3 + 3 = 6$, not 3. IV: $\frac{3}{2} + C = 2$ gives $C = \frac{1}{2}$.