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Separable Equations

  1. Aug 10, 2008 #1
    Hi I've been working on this problem repeatedly and thought I understood how to solve separable equations problems but I keep getting the wrong answer.

    y' = (1-2x)y^2

    Here's what I got for the problem:

    y'y^2 = (1-2x)

    [tex]\int(1\y^2) dy = \int(1-2x)dx[/tex]

    ln | y^2| = x-x^2 + C

    e^ln|y^2| =e^ x-x^2 +C , e^C = C

    y^2= + e^x-x^2 +C or y^2 = -e^x-x^2 +C

    y=[tex]\sqrt{}+ or - Ce^x-x^2[/tex]

    -1/6=+ or - [tex]\sqrt{}Ce^ 0-0^2[/tex]

    (-1\6)^2 = + or -([tex]\sqrt{}C[/tex])^2

    C = 1\36

    y=+ or - [tex]\sqrt{}1\36 e^x-x^2[/tex]

    Also, I'm not sure what to do when the problem tells you to determine the interval in which the solution is valid.

  2. jcsd
  3. Aug 11, 2008 #2


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    Homework Helper

    Welcome to PF!

    Hi Aero6! Welcome to PF! :smile:

    (have a squared: ² and an integral: ∫ :smile:)

    (and fractions in LaTeX are \frac{}{})

    erm … you got too excited at seeing the 1/ in the ∫ … they're not all logs! :wink:

    ∫dy/y² = … ? :smile:
  4. Aug 11, 2008 #3


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    Shouldn't it be y'/y^2= 1-2x ?

    Yes, it was!

    No, the integral of y-2 is -y-1

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