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Thread: 5dy/dx + 2x = 3.

  1. #1
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    5dy/dx + 2x = 3.

    The problem is $\displaystyle 5\frac{dy}{dx}$ +2x = 3

    by re-arranging I get the following

    y= $\displaystyle \frac{3}{5} -\frac{2x}{5}$ dx

    but need help with finding the general solution

    i integrated an came up with.

    y = $\displaystyle \frac{3^2}{6} - \frac{2x^2}{6}$ +C



    p.s. what is the source code for integral sign?
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  2. #2
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    \int is $\displaystyle \int.$

    You should write your rearrangement as

    $\displaystyle dy=\left(\dfrac{3}{5}-\dfrac{2x}{5}\right)dx,$ and then integrate. I'm not sure I buy your integration. Try again.
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  3. #3
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    okey i have tried again...

    $\displaystyle dy=\left(\dfrac{3x}{5}-\dfrac{?}{?}\right)dx$,

    but cannot see where this 2nd part comes from.
    My answer
    $\displaystyle dy=\left(\dfrac{3x}{5}-\dfrac{2x^2}{7}\right)dx$
    is different from the solution in the book. The solution in the book is

    $\displaystyle dy=\left(\dfrac{3x}{5}-\dfrac{x^2}{5}\right)dx$

    can you show me where im goin wrong please. many thanks
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    Where did the 7 come from?
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  5. #5
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    Quote Originally Posted by decoy808 View Post
    okey i have tried again...

    $\displaystyle dy=\left(\dfrac{3x}{5}-\dfrac{?}{?}\right)dx$,
    Why the questions marks? You had before
    $\displaystyle dy= \left(\dfrac{3}{5}- \dfrac{2x}{5}\right)dx$
    (except that you had "y" instead of "dy")
    Now integrate both sides: the integral of a constant is that constant times x and the integral of x is $\displaystyle (1/2)x^2$

    but cannot see where this 2nd part comes from.
    My answer
    $\displaystyle dy=\left(\dfrac{3x}{5}-\dfrac{2x^2}{7}\right)dx$
    $\displaystyle \dfrac{3x}{5}$ is correct. The integral of $\displaystyle \frac{2}{5}x$ is $\displaystyle \left(\frac{2}{5}\right)\left(\frac{1}{2}x^2\right )= \frac{1}{5}x^2$
    Perhaps you misread the $\displaystyle \frac{2}{5}\cdot\frac{1}{2}$ as $\displaystyle \frac{2}{5+ 2}$.

    is different from the solution in the book. The solution in the book is

    $\displaystyle dy=\left(\dfrac{3x}{5}-\dfrac{x^2}{5}\right)dx$

    can you show me where im goin wrong please. many thanks
    Last edited by HallsofIvy; Apr 6th 2011 at 12:03 PM.
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    need to go back over integration. thanks guys
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  7. #7
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    You're welcome for my contribution.
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