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Thread: linear ODE: quick question on equivalency

  1. #1
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    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:

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

    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!
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  2. #2
    MHF Contributor

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    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}$.
    Last edited by HallsofIvy; Mar 31st 2018 at 12:43 PM.
    Thanks from blaisem and topsquark
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  3. #3
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    Re: linear ODE: quick question on equivalency

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