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Math Help - need help witth this diff. eq.

  1. #1
    Junior Member
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    need help witth this diff. eq.

    y' + xy = x/y

    work done:

    y = vx => y' = v +v'x

    v +v'x + v(x^2) = 1/v =>

    v^2 + v'vx + (vx)^2 = 1 => v'vx + (v^2)(1+x^2) = 1

    ... it tappers off here i 'm afraid basically i thought i could turn it into a Bernouli eq. but i can't...

    i'v tried a few other substitutions bu i am hitting a wall,

    just a clue please
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  2. #2
    Super Member
    Joined
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    Quote Originally Posted by pepsi View Post
    y' + xy = x/y

    work done:

    y = vx => y' = v +v'x

    v +v'x + v(x^2) = 1/v =>

    v^2 + v'vx + (vx)^2 = 1 => v'vx + (v^2)(1+x^2) = 1

    ... it tappers off here i 'm afraid basically i thought i could turn it into a Bernouli eq. but i can't...

    i'v tried a few other substitutions bu i am hitting a wall,

    just a clue please

    Does this question require you to apply the Bernouli's method ?

    If not , I think we should use an other way : separating variables


     y'  + xy = \frac{x}{y}

     y' = \frac{dy}{dx} = x \left( \frac{1}{y} - y \right)


    by separating the variables ,

     \frac{y dy}{ 1-y^2 } = x dx

    integrating ,

      \frac{\ln( 1- y^2)}{2}= \frac{-x^2 + K }{2}

     \ln(1 - y^2  ) =- x^2 + K

     y = \sqrt{ 1 - e^{-x^2 + K }}

    Write  e^K  = c

     y = \sqrt{ 1 - ce^{-x^2}}
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  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
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    Thumbs up

    Quote Originally Posted by simplependulum View Post
    Does this question require you to apply the Bernouli's method ?

    If not , I think we should use an other way : separating variables


     y' + xy = \frac{x}{y}

     y' = \frac{dy}{dx} = x \left( \frac{1}{y} - y \right)


    by separating the variables ,

     \frac{y dy}{ 1-y^2 } = x dx

    integrating ,

     \frac{\ln( 1- y^2)}{2}= \frac{-x^2 + K }{2}

     \ln(1 - y^2 ) =- x^2 + K

     y = \sqrt{ 1 - e^{-x^2 + K }}

    Write  e^K = c

     y = \sqrt{ 1 - ce^{-x^2}}
    embarassed and thankful
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