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Thread: integrating factor

  1. #1
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    integrating factor

    is my integral R(x) and R(y) correct ? why it look so complicated ? integrating factor-0060.jpg
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    Re: integrating factor

    Neither are integrating factors because they are not expressions in a single variable. Can you post the full text of the question?
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    Re: integrating factor

    Quote Originally Posted by Archie View Post
    Neither are integrating factors because they are not expressions in a single variable. Can you post the full text of the question?
    the full text is Show that the first order differential equation is EXACT equation . (sin x - x sin y) dx + (cos y + y cos x ) dy = 0
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    Re: integrating factor

    Perhaps the differentials are transposed. The following is exact:
    $$(\sin x - x\sin y) \,\mathrm d y + (\cos y + y \cos x)\,\mathrm d x = 0 \\ \implies y\sin x + x\cos y = c$$
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    Re: integrating factor

    Quote Originally Posted by Archie View Post
    Perhaps the differentials are transposed. The following is exact:
    $$(\sin x - x\sin y) \,\mathrm d y + (\cos y + y \cos x)\,\mathrm d x = 0 \\ \implies y\sin x + x\cos y = c$$
    you mean the dy and dx should be interchanged , so that the equation is EXACT ?
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    Re: integrating factor

    That's not a valid operation. I'm suggesting that the question has been written incorrectly.
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    Re: integrating factor

    I will try and see
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    Re: integrating factor

    Quote Originally Posted by Archie View Post
    Perhaps the differentials are transposed. The following is exact:
    $$(\sin x - x\sin y) \,\mathrm d y + (\cos y + y \cos x)\,\mathrm d x = 0 \\ \implies y\sin x + x\cos y = c$$
    i tried this , but still didint get the answer . Is there any other alternative for this question?
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  9. #9
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    Re: integrating factor

    What answer are you looking for?
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    Re: integrating factor

    Quote Originally Posted by Archie View Post
    What answer are you looking for?
    the ans given is xcosy + ycosx = C
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  11. #11
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    Re: integrating factor

    So the exact equation would be
    $\displaystyle (\cos y - y\sin x) \, \mathrm d x + ( \cos x -x \sin y )\, \mathrm d y = 0$
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  12. #12
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    Re: integrating factor

    Quote Originally Posted by Archie View Post
    So the exact equation would be
    $\displaystyle (\cos y - y\sin x) \, \mathrm d x + ( \cos x -x \sin y )\, \mathrm d y = 0$
    sorry , i made a mistake here , the ans given is xcosy + ysinx = C , so what should the question look like ?
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  13. #13
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    Re: integrating factor

    See post #4.
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