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Find the exact value of a differential equation.

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    dy/dx= 200-2y. y(0)=75

    2. Relevant equations


    3. The attempt at a solution
    Do you move dx over and integrate.

    Do you just integrate it 200y-y^2+c
     
  2. jcsd
  3. Oct 19, 2016 #2

    andrewkirk

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  4. Oct 19, 2016 #3
    I'm having a little trouble with this. I move it so it becomes

    ##dy/dx=200-2y##
    ##dy=200dx-2ydx##
     
  5. Oct 19, 2016 #4

    Mark44

    Staff: Mentor

    No. This is completely wrong. I would strongly advise you to look at some examples of this technique in your textbook.

    Using separation of variables, you should end up with all terms involving y and dy on one side, and all terms involving x and dx on the other side.
    For this problem,
    1) Divide both sides by 200 - 2y
    2) Multiply both sides of the resulting equation by dx

    In this case, you should end up with ##\frac{dy}{200 - 2y} = dx##, or equivalently, ##\frac{dy}{100 - y} = 2 dx##,
     
  6. Oct 21, 2016 #5

    vela

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    I think what you're suggesting is to solve the problem as follows:
    \begin{align*}
    \frac{dy}{dx} &= 200-2y \\
    \int \frac{dy}{dx}\,dx &= \int (200-2y)\,dx \\
    y &= 200y-y^2+c
    \end{align*}
    There's a problem with that last step. If you don't recognize it, consider the questions below:

    Outside of this problem, if I asked you what ##\int 200\,dx## equalled, you'd hopefully say ##200 x+c##, yet in solving this problem, you encountered the exact same integral and (allegedly) said ##\int 200\,dx = 200y+c##.

    Similarly, consider the two integrals ##\int 2x\,dx## and ##\int 2y\,dx##. Why can you easily evaluate the first one but not the latter?
     
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