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Math Help - 5dy/dx + 2x = 3.

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

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

    by re-arranging I get the following

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

    but need help with finding the general solution

    i integrated an came up with.

    y = \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 \int.

    You should write your rearrangement as

    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...

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

    but cannot see where this 2nd part comes from.
    My answer
    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

    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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  4. #4
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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...

    dy=\left(\dfrac{3x}{5}-\dfrac{?}{?}\right)dx,
    Why the questions marks? You had before
    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 (1/2)x^2

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

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

    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; April 6th 2011 at 01:03 PM.
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  6. #6
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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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