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Thread: Understanding notation in a 1st order linear differentiation equation

  1. #1
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    Understanding notation in a 1st order linear differentiation equation

    Hi all,

    I'm having difficulty understanding conventions for DE, and I suppose calculus in general. I have not had calculus in several years, so I'm trying to figure out what exactly is happening in this sequence.

    Here's what the book says:
    $\displaystyle (du(t)/dt) = (1/2) (u(t)) $
    $\displaystyle (du(t)/dt) / u(t) = 1/2 $
    $\displaystyle (d/dt) ln|u(t)| = 1/2 $
    $\displaystyle ln |u(t)| = (1/2)t + C $
    $\displaystyle u(t) = ce^(t/2) $



    So, I understand the basics: rearrange the equation, integrate both sides, and solve for u(t).

    My only concern is that I have no idea what the derivative notation means. For example, what exactly is happening from line 2 to line 3 that makes the du(t) drop out. Does this mean that when I integrate (du(t)/dt) (1/ u(t)) that du drops out? Ok, but why is the (d/dt) left? Doesn't that mean "take the derivative"?

    $\displaystyle (d/dt) INTEGRAL((1/ u(t)) d(u) = 1/2$

    ^Is that what's happening? If so, then I still don't understand the next steps. Where is that d/dt going exactly??

    I hope you can understand what I'm asking: what do all these d's, dt's, and d(u)'s mean. I usually just ignore them and remember the process.

    Thanks.
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  2. #2
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    This is sloppy working as it skips steps.

    I'd do it like this...

    $\displaystyle \frac{du}{dt} = \frac{1}{2}u$

    $\displaystyle \frac{1}{u}\,\frac{du}{dt} = \frac{1}{2}$

    $\displaystyle \int{\frac{1}{u}\,\frac{du}{dt}\,dt} = \int{\frac{1}{2}\,dt}$

    $\displaystyle \int{\frac{1}{u}\,du} = \int{\frac{1}{2}\,dt}$

    $\displaystyle \ln{|u|} + C_1 = \frac{1}{2}t + C_2$

    $\displaystyle \ln{|u|} = \frac{1}{2}t + C_2 - C_1$

    $\displaystyle |u| = e^{\frac{1}{2}t + C_2 - C_1}$

    $\displaystyle |u| = e^{C_2 - C_1}e^{\frac{1}{2}t}$

    $\displaystyle u = \pm e^{C_2 - C_1}e^{\frac{1}{2}t}$

    $\displaystyle u = Ce^{\frac{1}{2}t}$ where $\displaystyle C = \pm e^{C_2 - C_1}$.
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  3. #3
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    Thanks for explaining in detail!

    On the third line, are the dt's crossing out? Is that what's actually happening?
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  4. #4
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    It's an application of the chain rule in reverse. It's almost like the $\displaystyle dt$s "cancel", but technically speaking you can't say that they cancel since a derivative is not a fraction, it only works like a fraction.
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