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Math Help - compute the general solution (1-x)y'=y^2

  1. #1
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    compute the general solution (1-x)y'=y^2

    compute the general solution (1-x)y'=y^2

    Can anyone, solve this step by step?
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  2. #2
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    Re: compute the general solution (1-x)y'=y^2

    Where did you get this problem? From a course in Calculus or Differential Equations? In either case you certainly should know how to integrate:
    [tex](1- x)y'= (1- x)\dfrac{dy}{dx}= y^2[tex]

    \dfrac{dy}{y^2}= y^{-2}dy= \frac{dx}{1- x}
    Can you integrate those?
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  3. #3
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    Re: compute the general solution (1-x)y'=y^2

    dy/(y^2)=dx/(1-x)
    y^-2dy=ln (1-x)+c
    -1/y=ln (1-x)+c

    I stock from here.
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  4. #4
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    Re: compute the general solution (1-x)y'=y^2

    It is ok just solve for y.......(make y the subject)
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  5. #5
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    Re: compute the general solution (1-x)y'=y^2

    One thing you can do is "flip" both sides:
    \frac{-1}{y}= ln|1- x|+ c becomes
    -y= \frac{1}{ln|1- x|+ c}

    (Notice the absolute value. ln(x) isn't defined for x\le 0. If you know x< 1, for example, if you given a value for y(0)= 0, you can use just "1- x". On the other hand, if you have reason to believe x>1, say you are given a value of y(1), you need "x- 1".)
    Last edited by HallsofIvy; October 26th 2013 at 09:05 AM.
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