# How do you find the general solution of the differential equation: dy/dx=(3x)/y?

Dec 31, 2016

y = +-sqrt(3x^2 + C, (where $C$ is arbitrary constant).

#### Explanation:

We have:

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

This is a First Order separable Differential Equation, so we can just collect terms in $y$, and terms in $x$ and "separate the variables" to get:

$\int \setminus y \setminus \mathrm{dy} = \int \setminus 3 x \setminus \mathrm{dx}$

We can now integrate to we get:

$\setminus \setminus \frac{1}{2} {y}^{2} = \frac{3}{2} {x}^{2} + C '$
$\therefore {y}^{2} = 3 {x}^{2} + C$, (where $C$ is arbitrary constant).
$\therefore y \setminus \setminus = \pm \sqrt{3 {x}^{2} + C}$, (where $C$ is arbitrary constant).