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Math Help - How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)*6

  1. #1
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    How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)*6

    How is:


    e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6


    the same as:


    (d/dx)*[e^(-3x)*y]=e^(-3x)*6
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  2. #2
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    Re: How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)

    Are you saying that you do not know how to take the derivative of e^{-3x}y with respect to x?
    Last edited by HallsofIvy; August 11th 2012 at 10:23 AM.
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  3. #3
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    Re: How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)

    No, I'm learning the method of solving linear 1st-order differential equations, and I have the example:

    dy/dx + 3y = 2xe^(-3x)

    I found e^(3x) to be the integrating factor, so then I multiplied it through the original equation giving me:

    e^(3x)*(dy/dx) + e^(3x)*3y = 2x

    this next step is what is confusing me, (and I realize that it is probably something very simple) I'm supposed to use the product rule where: u=e^(3x), dv=(dy/dx), du=e^(3x), & v=3y

    Then the resulting equation is:

    [e^(3x)*y]=2x, so my question is:

    Can you show me the individual steps for where:

    e^(3x)*(dy/dx) + e^(3x)*3y = 2x

    becomes:

    [e^(3x)*y]=2x

    Thank you.
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