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Math Help - Solve the following ODE ..#2

  1. #1
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    Solve the following ODE ..#2

    Hello:

    Problem:
    Solve the following equation :
    2y(x+y+2)dx+(y^2-x^2-4x-1)dy=0

    Solution:
    I did not get it ..
    not separable, not homogeneous, not exact, not linear, not bernoulli
    So I tried to use the method of determinating the integrating factor ..
    But this failed with me:

    \dfrac{1}{N} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right) is not a function in x only ..

    and


    \dfrac{1}{M} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right) is not a function in y only ..

    any ideas??
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  2. #2
    Super Member 11rdc11's Avatar
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    It is an exact equation

    \dfrac{1}{M} \left( \dfrac{\partial N}{\partial y} - \dfrac{ \partial M }{ \partial x} \right)


    \frac{(-2x-4)-(2x+4y+4)}{2xy+2y^2+4y} = \frac{-4x-4y-8}{2xy+2y^2+4y} = \frac{-x-y-2}{xy+y^2+2y} = \frac{-(x+y)-2}{y[(x+y)+2]} = \frac{-(x+y)-2}{-y[-(x+y)-2]}

    so

    it equals

    \frac{-1}{y}

    Can you finish up from here?
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  3. #3
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    Wait
    In my notes & book, its only :

    \dfrac{1}{N} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right) and it should be function in x only

    &

    \dfrac{1}{M} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right) and it should be function in y only

    There is no thing about : \dfrac{1}{M} \left( \dfrac{\partial N}{\partial y} - \dfrac{ \partial M }{ \partial x} \right)


    any one know is the latter valid also ??
    if so, can you tell me all expressions??
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Liverpool View Post
    Wait
    In my notes & book, its only :

    \dfrac{1}{N} \left( \dfrac{\partial M}{\partial y} - \dfrac{ \partial N }{ \partial x} \right) and it should be function in x only

    &

    \dfrac{1}{M} \left( \dfrac{ \partial N }{ \partial x}-\dfrac{\partial M}{\partial y} \right) and it should be function in y only
    those are the correct expressions.
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  5. #5
    Super Member 11rdc11's Avatar
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    sorry, I had just copied your latex and switched the M and N and forgot to switch the partial dy and dx when correcting the error. The rest of the work is correct though.
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  6. #6
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    Thanks.
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