# What is a solution to the differential equation dy/dx= y^2? (Please)

Apr 17, 2018

$y \left(x\right) = - \frac{1}{x + C}$

#### Explanation:

This is a separable differential equation. We'll get all $y , \mathrm{dy}$ on the right side and all $x , \mathrm{dx}$ on the left.

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2}$

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

Note that we have no $x$ so all we do is move $\mathrm{dx}$ to the left.

${y}^{-} 2 \mathrm{dy} = \mathrm{dx}$

Integrate both sides:

$\int {y}^{-} 2 \mathrm{dy} = \int \mathrm{dx}$

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

We need an explicit solution in the form $y \left(x\right) :$

$- 1 = y \left(x + C\right)$

$y \left(x\right) = - \frac{1}{x + C}$

Apr 17, 2018

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

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2}$
${y}^{- 2} \mathrm{dy} = \mathrm{dx}$
$\int {y}^{- 2} \mathrm{dy} = \int \mathrm{dx}$
$- {y}^{- 1} = x + C$
${y}^{- 1} = - x - C$
$y = \frac{1}{- x - C}$
$y = - \frac{1}{x + C}$, $C$ is a constant