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Math Help - finding a value to make a DE exact

  1. #1
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    finding a value to make a DE exact

    Please help me out!!!

    this is the question: Find the value for k so that the given DE is exact

    (6x (y^3) + cos y )dx + (2k (x^2) (y^2) - x sin y)dy =0


    thank you so much
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kithy View Post
    Please help me out!!!

    this is the question: Find the value for k so that the given DE is exact

    (6x (y^3) + cos y )dx + (2k (x^2) (y^2) - x sin y)dy =0


    thank you so much
    For a DE to be exact, the following must be true:

    \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}

    From your DE, we see that:

    M(x,y)=6xy^3+\cos y\implies \frac{\partial M}{\partial y}=18xy^2-\sin y

    N(x,y)=2kx^2y^2-x\sin y\implies \frac{\partial N}{\partial x}=4kxy^2-\sin y

    Thus,

    18xy^2-\sin y=4kxy^2-\sin y

    Try taking it from here.

    Hope that this makes sense!

    --Chris
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  3. #3
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    Ok, i equated the two equations and i came out with K= (18xy^2)/(4xy^2) and so i am assuming the xy^2 cancel out and we are left with k=9/2 is that right?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kithy View Post
    Ok, i equated the two equations and i came out with K= (18xy^2)/(4xy^2) and so i am assuming the xy^2 cancel out and we are left with k=9/2 is that right?
    Sorry for the late response. I've been in school all day...

    Yes, that is correct. Does is make more sense now?

    --Chris
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  5. #5
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    Yes it really does, thank you so much, i understand it better now
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