Question

Solve the differential equation.

dy/dx = xy^2

Answer 1

You can solve this differential equation by separating the variables and differentiating both sides.

1) Multiply both sides by dx and divide both sides by $y^{2}$.
$\frac{dy}{dx} = xy^2\\ \frac{1}{y^2} \: dy = x\:dx$

2) Integrate both sides. Remember the power rule for integrals. Say you have a value $x^{n}$, where $n \neq -1$. Take the power, n, and add 1. Then divide the new expression $x^{n+1}$ by the new power, n + 1. The integral of $x^{n}$ would be $\frac{x^{n+1}}{n+1}$ (+C, if it is an indefinite integral). Remember that you can subtract C from both sides and just have C on one side (since the constant doesn't have a definite value):
$\frac{1}{y^2} \: dy = x\:dx\\ \int \frac{1}{y^2} \: dy= \int x\:dx\\ \int y^{-2} \: dy= \int x\:dx \\ \frac{ y^{-2+1}}{-2+1} + C = \frac{x^{1+1}}{1+1} + C\\ - y^{-1} = \frac{ x^{2}}{2} + C\\ -\frac{2}{y} = x^{2} + C\\ y = - \frac{2}{x^{2} + C}$

Your solution is $y = - \frac{2}{x^{2} + C}$.

Answer 2

$\frac{dy}{dx} = xy^2\ \Rightarrow\ \frac{dy}{y^2} = x \;dx\ [y \ne 0] \ \Rightarrow\ \int y^{-2} dy = \int x \, dx \Rightarrow \\ \\ -y^{-1} = \frac{1}{2}x^2 + C\ \Rightarrow\ \frac{1}{y} = -\frac{1}{2}x^2 - C \Rightarrow \\ \\ y = \displaystyle\frac{1}{-\frac{1}{2}x^2 - C} = \frac{2}{K - x^2},\ \text{where K = -2C }$

y = 0 is also a solution.