Integrating Factor

Problem:

$ \frac{2y}{x^2}dx + \frac{1}{x}dy = 0 $.

Solution:

Let $ M(x, y) = \frac{2y}{x^2} $, $ N(x, y) = \frac{1}{x} $, now we have

$ \frac{\partial M}{\partial y} = \frac{2}{x^2} \ne \frac{-1}{x^2} = \frac{\partial N}{\partial x} $

But they look similar, so let's assume there exists $ \mu(x) $ such that

$ \frac{\partial \mu M}{\partial y} = \frac{\mu \partial N}{\partial x} $

Expanding, we get

$ \mu \frac{\partial M}{\partial y} = \frac{\partial \mu}{\partial x} N + \mu \frac{\partial N}{\partial x} $

or

$ \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})= \frac{1}{\mu}\frac{\partial \mu}{\partial x}$

so

$ \mu = e^{\int \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) dx} $

Putting in the values, we have

$ \begin{eqnarray*} \mu &=& e^{\int \frac{1}{\frac{1}{x}}(\frac{2}{x^2} - \frac{-1}{x^2}) dx} \\ &=& e^{\int \frac{1}{\frac{3}{x}} dx} \\ &=& e^{3 \log x} \\ &=& x^3 \end{eqnarray*} $

The rest is rather easy now, we need to find $ f(x, y) $ such that $ \frac{\partial f}{\partial x} = x^3 \frac{2y}{x^2} = 2xy $ and $ \frac{\partial f}{\partial y} = x^3 \frac{1}{x} = x^2 $. We found $ f(x, y) = x^2y = c $, therefore the solution to the differential equation is $ y = cx^{-2} $.