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Math Help - Initial Value Differential Equation

  1. #1
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    Initial Value Differential Equation

    I have dy/dx = x^2/y^2

    =y^2 dy = x^2 dx

    y^2 dy = x^2 dx

    y^3/3 = x^3/3+ C

    y^3 = x^3 + 3C

    3C= some constant k

    y = (x^3+k)^(1/3)

    I now have to solve the initial value problem of y(0)=2 ; kinda forgot how to do this part.

    Do I just set it equal to 2 then plug in 0 so k=8 then do i multiply by 3 ?
    Last edited by goku900; March 26th 2013 at 12:10 PM.
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  2. #2
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    Re: Initial Value Differential Equation

    Just set x = 0 and y = 2

    2 = (0 + k)^(1/3)
    2 = k^(1/3)

    cube both sides

    8 = k
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