# Thread: linear ODE: quick question on equivalency

1. ## linear ODE: quick question on equivalency

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:

$\displaystyle \label{eq:some} \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=1$

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:

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

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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!

2. ## Re: linear ODE: quick question on equivalency

The problem is that you are not differentiating correctly!

To differentiate $\frac{d\frac{1}{x}y}{dx}$, since both $\frac{1}{x}$ and $y$ are functions of x, use the product rule.

$\frac{d\frac{1}{x}y}{dx}= $$\left(\frac{d\frac{1}{x}}{dx}\right)y+$$ \frac{1}{x}\frac{dy}{dx}$.

3. ## Re: linear ODE: quick question on equivalency

Woops forgot y was a function of x. Thank you!