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Math Help - Seperable Differential Equation

  1. #1
    Member Marconis's Avatar
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    Seperable Differential Equation

    We covered this last Tuesday, and had an exam on Thursday, so instead of doing HW I studied for the exam. I am a little confused on DE, and forgot a bit of what we went over in class. Could you check my work, please?
    <br />
\frac{dy}{dx}=\frac{\sqrt{x}}{e^y}

    e^y dy= \sqrt{x}dx

    \int e^y dy= \int \sqrt{x}dx

    e^y= \frac{2}{3}x^\frac{3}{2} + C

    ylne=\frac{2}{3}x^\frac{3}{2} + C

    y=ln(\frac{2}{3}x^\frac{3}{2} + C)

    Thank you.
    Last edited by mr fantastic; April 10th 2011 at 09:11 PM. Reason: Moved post as requested by user (accidentally posted in wrong subforum); no infraction given.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Marconis View Post
    We covered this last Tuesday, and had an exam on Thursday, so instead of doing HW I studied for the exam. I am a little confused on DE, and forgot a bit of what we went over in class. Could you check my work, please?
    <br />
\frac{dy}{dx}=\frac{\sqrt{x}}{e^y}

    e^y dy= \sqrt{x}dx

    \int e^y dy= \int \sqrt{x}dx

    e^y= \frac{2}{3}x^\frac{3}{2} + C

    ylne=\frac{2}{3}x^\frac{3}{2} + C

    y=ln(\frac{2}{3}x^\frac{3}{2} + C)

    Thank you.
    You're pretty much correct, but note the minor thing you missed:

    y\ln e = \frac{2}{3}x^{3/2}+C should really be y\ln e = \ln(\frac{2}{3}x^{3/2}+C).

    When you take ln of one side, you have to do the same for the other.

    Besides that, I don't see anything wrong with your solution.
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  3. #3
    Member Marconis's Avatar
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    Those are the kinds of silly mistakes that tend to cost me on exams. Thanks for pointing that out, and thank you for the speedy reply.
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