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March 26, 2017

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dy/dx = 4xy^2
I need to find the integral of that. So I separate the variables and have 1/y^2 dy = 4x dx. Where do I go from here?

  • Calculus - ,

    y^-2 dy = 4 x dx

    -1 y^-1 + c = 2 x^2 the important step

    -1/y + c = 2 x^2

    -1 + c y = 2 x^2 y
    y (2 x^2 -c) = -1
    y = -1/ (2 x^2 -c) you can write that several ways

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