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Help with a simple initial value problem

  1. Jan 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the IVP dy/dt=4y-2, y(0)=3


    This is how I work it. I integrate both sides and get y= 2y^2 - 2y + C

    I then solve for C: 3=2(0)^2 - 2(0) + C

    C = 3

    Is this the correct way to solve for C?
     
  2. jcsd
  3. Jan 8, 2009 #2

    rock.freak667

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    Homework Helper

    [tex]\frac{dy}{dt}=4y-2[/tex]

    [tex] \frac{1}{4y-2}dy= 1 dt[/tex]

    Separation of variables. Now integrate.
     
  4. Jan 8, 2009 #3

    HallsofIvy

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    It's not solving for C that is the problem! You have integrated on the left side with respect to t and on the right with respect to y. You can't do that. In order to integrate with respect to t you have to "multiply" by dt: (dy/dt) dt= dy, but then the right side is (4y- 2)dt which is NOT the same as (4y- 2)dy.

    What you can do, as rock.freak667 suggested is "separate" the equation so you have only y on one side and t on the other:
    dy= (4y-2)dt so dy/(4y-2)= dt. NOW you can integrate the left side with respect to y (because of the "dy") and the right side with respect to t (because of the "dt"). After you have done that you can set y= 3, t= 0 to find C, the constant of integration.
     
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