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Math Help - 3. What method I can use on this equation?

  1. #1
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    3. What method I can use on this equation?

    It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

    Question
     (x^3 + y^3) dx + y^2(3x + ky) dy = 0 ;  k is a constant

    Answer
     ky^4 + 4xy^3 + x^4 = c
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  2. #2
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    Quote Originally Posted by redfox2600 View Post
    It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

    Question
     (x^3 + y^3) dx + y^2(3x + ky) dy = 0 ;  k is a constant

    Answer
     ky^4 + 4xy^3 + x^4 = c
    It's similar to others you have asked. Divide through by x^3 and re-arrange. What do you notice?

    If you need more help, please post your work.
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  3. #3
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    Is this correct? correct if I am wrong.
    I used the exact method.



    Question
     (x^3 + y^3) dx + y^2(3x + ky) dy = 0

    Answer
     ky^4 + 4xy^3 + x^4 = c


    Solution:
     (x^3 + y^3) dx + (3xy^2 + ky^3) dy = 0

     M = (x^3 + y^3)

     N = (3xy^2 + ky^3)

     \frac {\partial M}{\partial y} = 3y^2

     \frac {\partial N}{\partial x} = 3y^2

     \frac {x^4}{4} + xy^3 + \frac {ky^4}{4} = c

     ky^4 + 4xy^3 + x^4 = c


    thanks mr fantastic.
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  4. #4
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    Yes, that is correct.
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