# What is the solution to the Differential Equation 4/y^3 dy/dx=1/x?

Jun 12, 2017

$y = \pm \sqrt{\frac{2}{C - \ln | x |}}$

#### Explanation:

We have:

$\frac{4}{y} ^ 3 \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x}$

This is a first Order Separable Differential Equation, we can just "separate the variables" to get

$\int \setminus \frac{4}{y} ^ 3 \setminus \mathrm{dy} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$

And integrating gives us:

$4 \setminus {y}^{- 2} / \left(- 2\right) = \ln | x | + C$

$\therefore - \frac{2}{y} ^ 2 = \ln | x | + C$

$\therefore {y}^{2} = - \frac{2}{\ln | x | + C}$

$\therefore {y}^{2} = \frac{2}{C - \ln | x |}$

$\therefore y = \pm \sqrt{\frac{2}{C - \ln | x |}}$