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Thread: integral of (1/y)(dy/dx) dx

  1. #1
    Sep 2016

    integral of (1/y)(dy/dx) dx

    Hi there,

    I'm working through a textbook section on the separation of variables method with respect to differentials and I want to be sure that I understand the why/how/etc. behind the simplification below.
    integral of (1/y)(dy/dx) dx-integral-problem.png
    Do the dxs just cancel each other out like any simple x, y, etc. variable would? But don't the two dxs represent different things (one represents the variable to differentiate with respect to and the other the variable to integrate with respect to)? Or doesn't that matter? If someone could just provide an explanation I would really appreciate it!

    Sorry if I'm asking a stupid question!


    Attached Thumbnails Attached Thumbnails integral of (1/y)(dy/dx) dx-integral-problem.png   integral of (1/y)(dy/dx) dx-integral-problem.png  
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  2. #2
    MHF Contributor

    Apr 2005

    Re: integral of (1/y)(dy/dx) dx

    Strictly speaking it is not just a matter of "canceling the dx s", it is an application of the chain rule. If y is a function of x, then $\displaystyle \frac{d(f(y)}{dx}= \frac{df}{dy}\frac{dy}{dx}$.

    In differential form, that is $\displaystyle \frac{df}{dy}\frac{dy}{dx}dx= \frac{df}{dy}dy$. In particular, with $\displaystyle f(y)= ln(y)$, $\displaystyle \frac{df}{dy}= \frac{1}{y}$ so that $\displaystyle \frac{d(ln(y))}{dy}= \frac{d(ln(y))}{dy}dx= \frac{1}{y}dx$ and $\displaystyle \int \frac{1}{y}dy= \int\frac{d(ln(y)}{dy}\frac{dy}{dx}dx= \int \frac{d(ln(y)}{dy}dy= ln(y)+ C$.
    Last edited by skeeter; Sep 14th 2016 at 11:50 AM. Reason: fix latex tag
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