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Math Help - Stumped

  1. #1
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    Stumped

    "Verify that the given differential equation is exact; then solve it."

    (e^(x) * siny + tany)dx + (e^(x) * cosy + x * (secy)^2)dy = 0
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by johnnymath345 View Post
    "Verify that the given differential equation is exact; then solve it."

    (e^(x) * siny + tany)dx + (e^(x) * cosy + x * (secy)^2)dy = 0
    To be exact, the term in front of dx must be f_x
    and the term in front of dy must be f_y, WHERE the bottom line is to find that f(x,y).
    Ignore the silly N and M in must books, it's foolish.
    Next from Clarabell's theorem f_{xy}=f_{yx}
    and in our case that's correct, we get e^x\cos y in both cases.
    So, it's exact. Hence integrate the f_x wrt x
    and f_y wrt y to get this f(x,y) and when you integrate, don't just add +C
    in the first case it's g(y) and the second it's h(x).
    Last edited by matheagle; February 26th 2009 at 08:56 AM.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    To be exact, the term in front of dx must be f_x
    and the term in front of dy must be f_y, WHERE the bottom line is to find that f(x,y).
    Ignore the silly N and M in must books, it's foolish.
    Next from Clarabell's theorem f_xy=f_yx
    You mean f_{xy}= f_{yx}
    when you have more than one symbol after _ or ^, you need { } around them.

    and in our case that's correct, we get e^x\cos y in both cases.
    So, it's exact. Hence integrate the f_x wrt x
    and f_y wrt y to get this f(x,y) and when you integratse, don't just add +C
    in the first case it's g(y) and the second it's h(x).
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  4. #4
    MHF Contributor matheagle's Avatar
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    eye no, it was late and I was too tired to check, thanks
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