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Help with DE problem and seperation of variables

  1. Apr 30, 2005 #1
    i found the DE for a problem, and it was y'(t) = .06 - y/1040

    and then the problem gave me the following hint:

    Find the amount of sugar after t minutes. Note: When you solve by separation of variables, keep the coefficient of y on the right side and bring over to the denominator on the left side an expression of the form (y - some constant).

    what does this mean???? can anyone actually show me how to do this?? thanks
     
  2. jcsd
  3. Apr 30, 2005 #2

    AKG

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    EDIT: Of course, you can do it with separation of variables, I just wasn't thinking. See StatusX's post below.

    1) Take the derivative again, so y'' = (-1/1040)y'. You get y' = Ce(-t/1040), and you can integrate this to find y = De(-t/1040) + E, and then plug this back into the original DE and solve for D in terms of E, or vice versa. You will be left with one unknown, since no initial conditions are specified.

    2) Find the solution to the associated homogeneous equation, y'(t) = -y/1040, let's call that solution ch(t), where c is a co-efficient that can vary (i.e. ch(t) will be a solution to the homogeneous equation for any real c). Then, find any particular solution p(t). The constant function p(t) = 1040/0.06 will do the trick. Your answer will be y = ch + p.

    This is generally true. If you have an n-th order differential equation (so it includes the n-th derivative, whereas this is a 1st order equation since it only includes up to the 1st derivative), you will have n linearly independent homogeneous solutions, h1, h2, ..., hn. Then, if you know any particular solution p, the general solution will be:

    c1h1 + c2h2 + ... + cnhn + p

    for any choice of c1, ..., cn. This is obviously a family of solutions, but if you look at all possible choices of c1, ..., cn, then this will give you all possible solutions.

    3) Look at the equation as y'(t) + a(t)y(t) = b(t), where a(t) = 1/1040 and b(t) = 0.6. Your solution will be:

    [tex]\frac{\int u(x)b(x)\, dx + C}{u(x)}[/tex]*

    where

    [tex]u(x) = \exp \left (\int a(x)\, dx\right )[/tex]

    This is also true in general, if you have an equation of the form y' + ay = b. All solutions will be defined by * for all choices of C.
     
    Last edited: Apr 30, 2005
  4. Apr 30, 2005 #3

    StatusX

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    [tex]dy/dt = 0.06 - y/1040 [/tex]

    [tex] \frac{dy}{0.06 - y/1040} = dt [/tex]

    [tex] \int \frac{dy}{0.06 - y/1040} = \int dt [/tex]

    [tex] -1040 ln(0.06 - y/1040) = t + C [/tex]

    and so on
     
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