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Math Help - [SOLVED] Problem

  1. #1
    Member Altair's Avatar
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    [SOLVED] Problem

    The following is an Exact Differential Equation :-

    (\frac{x+y}{y-1}) dx -\frac{1}{2}(\frac{x+1}{y-1})^2dy

    Multiplied both sides with 2(y-1)^2.

    Then opened -(x+1)^2 and (y-1)(x+y).

    After partial differentiation the two sides doesn't come the same whereas this whole exercise is composed of questions which are Exact. And the end answer also doesn't come correct.

    Is there a problem with my approach or what ?
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  2. #2
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    Hello, Altair!

    Check your work . . .


    The following is an Exact Differential Equation:

    . . \left(\frac{x+y}{y-1}\right) dx -\frac{1}{2}\left(\frac{x+1}{y-1}\right)^2dy \;=\;0

    We have: . M \;=\;\frac{x+y}{y-1}

    . . Then: . M_y \:=\:\frac{(y-1)\!\cdot\!1 - (x+y)\!\cdot\!1}{(y-1)^2} \;=\;\boxed{-\frac{x+1}{(y-1)^2}}


    We have: . N \;=\;-\frac{1}{2}\frac{(x+1)^2}{(y-1)^2} \;=\;-\frac{1}{2(y-1)^2}(x+1)^2

    . . Then: . N_x\;=\;-\frac{1}{2(y-1)^2}\cdot2(x+1) \;=\;\boxed{-\frac{x+1}{(y-1)^2}}

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