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 $y = cx^{-2}$.