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Math Help - seperating variables

  1. #1
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    seperating variables

    If dy/dx = -10y and if y = 50 when x = 0, then y = (a) 50cos10x (b) 50e^(10x) (c) 50e^(-10x) (d) 50 - 10x (e) 50 - 5x^2
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  2. #2
    Jen
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    Quote Originally Posted by LeoBloom. View Post
    If dy/dx = -10y and if y = 50 when x = 0, then y = (a) 50cos10x (b) 50e^(10x) (c) 50e^(-10x) (d) 50 - 10x (e) 50 - 5x^2

    Notice this is a seperable equation

    \frac{dy}{dx}=-10y

    Getting all of the y's on one side gives

    \frac{1}{y}dy=-10dx

    Now you integrate both sides


    \ln|y|=-10x+C

    Now solving for y

    y=e^{-10x+C}=e^{-10x}e^{C}=ke^{-10x} where k=\pm e^{C}

    Now using the intial condition you can solve for k.

    50=ke^{-10(0)}

    50=k

    So the solution to the differential equation is

    y=50e^{-10x}
    Last edited by Jen; May 30th 2008 at 10:31 PM.
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  3. #3
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    Have you come across the technique of separating variables yet?

    for your problem we have
    <br />
\frac{dy}{dx}=-10y

    divide through by y and integrate both sides with respect to x
    <br />
\int \frac{1}{y} \frac{dy}{dx} dx=\int -10 dx

    giving us
    <br />
\int \frac{1}{y}dy=\int -10 dx

    do the integration leaving us with
    ln |y| = -10x+c

    I'll leave you to finish it off and plug in the boundary condition
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  4. #4
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    Right, thanks alot guys, yes, we have done this before in class but for some reason it flew out of my head.
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