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  1. #1
    Member Altair's Avatar
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    Help required...

    Q. $\displaystyle (2x + y)dy - (4x +2y -1)dy = 0$

    After using $\displaystyle Z = 2x +y$, I got the answer,

    $\displaystyle c = 2y -ln(2x + y)$

    However, its very different from the answer at the back of the book. I have checked and rechecked but ended up with the same wrong answer.
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  2. #2
    Flow Master
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    Quote Originally Posted by Altair View Post
    Q. $\displaystyle (2x + y){\color{red}dy} - (4x +2y -1){\color{blue}dy} = 0$

    Mr F says: You have two dy's here. I'm guessing one of them is meant to be a dx. Which one - the red or the blue?

    After using $\displaystyle Z = 2x +y$, I got the answer,

    $\displaystyle c = 2y -ln(2x + y)$

    However, its very different from the answer at the back of the book. I have checked and rechecked but ended up with the same wrong answer.
    ..
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  3. #3
    Member Altair's Avatar
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    The red one.
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  4. #4
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    Quote Originally Posted by Altair View Post
    The red one.
    $\displaystyle \frac{dy}{dx} = \frac{2x + y}{4x + 2y - 1}$.

    Substitute $\displaystyle z = 2x + y \Rightarrow y = z - 2x$. Then $\displaystyle \frac{dy}{dx} = \frac{dz}{dx} - 2$.

    Therefore:

    $\displaystyle \frac{dz}{dx} - 2 = \frac{z}{2z - 1}$

    $\displaystyle \frac{dz}{dx} = \frac{z}{2z - 1} + 2 = \frac{5z - 2}{2z - 1}$


    $\displaystyle \Rightarrow \frac{dx}{dz} = \frac{2z - 1}{5z - 2} = \frac{1}{5} \, \left( 2 - \frac{1}{5z - 2}\right)$.


    Therefore:

    $\displaystyle x = \frac{1}{5} \, \left( 2z - \frac{1}{5} \ln |5z - 2| + C \right) $


    $\displaystyle \Rightarrow 5x = 2z - \frac{1}{5} \ln |5z - 2| + C$


    $\displaystyle \Rightarrow 5x = 2(2x + y) - \frac{1}{5} \ln |5(2x + y) - 2| + C$.


    I'll leave to you to take it as further as necessary.
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