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Math Help - Find general solution for given points

  1. #1
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    Find general solution for given points

    Thanks to all for your help! Please show me where i went wrong. I'm getting zeros as the answer for part a) and b)

    Given 2xy(dy/dx) + y^2 - 1 = 0 find all solutions A) (1,0) B) (0,1) in the x-y plane.

    Here's my workout:
    2xy (dy/dx) +y^2 -1 = 0
    (2xy/1-y^2) (dy/dx) =1
    (2y/1-y^2) (dy/dx) =1/x

    integrate both sides:
    S (2y/(1-y^2) (dy/dx) = S 1/x dx
    -lnly^2 - 1ldy = ln lxl + C

    solve for y get rid of ln take e to both sides
    e^(-lnly^2 - 1) = e^(ln lxl + C)

    General solution:
    y= square root x(e^(C/2))
    Last edited by nivek0078; September 30th 2012 at 06:22 PM.
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  2. #2
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    Re: Find general solution for given points

    Quote Originally Posted by nivek0078 View Post
    Thanks to all for your help! Please show me where i went wrong. I'm getting zeros as the answer for part a) and b)

    Given 2xy(dy/dx) + y^2 - 1 = 0 find all solutions A) (1,0) B) (0,1) in the x-t plane.

    Here's my workout:
    2xy (dy/dx) +y^2 -1 = 0
    (2xy/1-y^2) (dy/dx) =1
    (2y/1-y^2) (dy/dx) =1/x

    integrate both sides:
    S (2y/(1-y^2) (dy/dx) = S 1/x dx
    -lnly^2 - 1ldy = ln lxl + C

    solve for y get rid of ln take e to both sides
    e^(-lnly^2 - 1) = e^(ln lxl + C)

    General solution:
    y= square root x(e^(C/2))
    The x-t plane? Why would you be in the x-t plane when your DE is in terms of x and y?
    Thanks from nivek0078
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  3. #3
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    Re: Find general solution for given points

    I'm sorry it is the x-y plane!
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  4. #4
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    Re: Find general solution for given points

    First of all, when you integrate with the substitution on the LHS, you should end up with \displaystyle \begin{align*} -\ln{\left| 1 - y^2 \right|} \end{align*}
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  5. #5
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    Re: Find general solution for given points

    Yes I have that. When i plug in the points it doesn't generate the right soltuion. can't seem to find where i went wrong in the process.

    integrate both sides:
    S (2y/(1-y^2) (dy/dx) = S 1/x dx

    -ln ly^2 - 1 ldy = ln lxl + C

    Then take e to both sides
    e^(-lnly^2 - 1) = e^(ln lxl + C)

    General solution
    y= square root x(e^(C/2))
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  6. #6
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    Re: Find general solution for given points

    Quote Originally Posted by nivek0078 View Post
    Yes I have that. When i plug in the points it doesn't generate the right soltuion. can't seem to find where i went wrong in the process.

    integrate both sides:
    S (2y/(1-y^2) (dy/dx) = S 1/x dx

    -ln ly^2 - 1 ldy = ln lxl + C

    Then take e to both sides
    e^(-lnly^2 - 1) = e^(ln lxl + C)

    General solution
    y= square root x(e^(C/2))
    I am quite capable of reading what you wrote. You, however, need to make note of the mistake you made which I pointed out. You have just rewritten exactly the same thing!
    Thanks from nivek0078
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  7. #7
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    Re: Find general solution for given points

    Sorry didn't see it right off. Thank you for the help.
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