Was reading how to solve a linear ODE and found this site with a method that seems straightforward to use.

Problem is I didn't understand steps 4-5 toward the bottom, namely the equation from step 4:

$\displaystyle \begin{equation} \label{eq:some}

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

\end{equation}$

is equivalent to the one in step 5:

$\displaystyle \frac{d}{dx}(\frac{1}{x}y)=1 $

or putting them together for conciseness, the following equivalency:

$\displaystyle \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=\frac{d}{dx}(\frac{1}{x}y)$

-----

My thought process is that if I follow through with the derivative on the right-hand side, I obtain

$\displaystyle \frac{d}{dx}(\frac{1}{x}y)=-\frac{1}{x^2}y$

substituting this into the equivalency above, I get

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

$\displaystyle \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y+(\frac{1}{x^2}y)=-\frac{1}{x^2}y+(\frac{1}{x^2}y)$

$\displaystyle \frac{1}{x}\frac{dy}{dx}=0$

but is this correct it's equal to zero?

Thank you!