# What is a solution to the differential equation xy' + 2y = 0?

Sep 8, 2016

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

#### Explanation:

Write $y '$ as $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$x \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y = 0$

Try to separate the variables:

$x \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 y$

$- \frac{1}{2 y} \mathrm{dy} = \frac{1}{x} \mathrm{dx}$

Integrate both sides:

$- \frac{1}{2} \int \frac{1}{y} \mathrm{dy} = \int \frac{1}{x} \mathrm{dx}$

$- \frac{1}{2} \ln y = \ln x + C$

Remember that any modification to $C$ is basically immaterial, it stays $C$ because it becomes some other unimportant constant:

$\ln y = - 2 \ln x + C$

y=e^(-2lnx+C

$y = {e}^{- 2 \ln x} \cdot {e}^{C}$

$y = C {e}^{\ln \left({x}^{-} 2\right)}$

$y = C {x}^{-} 2$

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