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Help with initial value problem

  1. Jan 13, 2009 #1
    Solve: dy/dt = 4y-2, y(0)=3

    I first seperate variables: dy/(4y-2) = dt

    I then integrate both sides to get: ln|4y-2|=t + c

    I next set 4y-2=Ce^t

    Do I just plug 3 in for y to get C=10 or am I going about this problem the wrong way??
     
  2. jcsd
  3. Jan 13, 2009 #2

    Mark44

    Staff: Mentor

    You're close, but made an error.
    Your antiderivate for 1/(4y - 2) is not ln|4y - 2|.

    Factor 4 from the two terms in the denominator, and then multiply both sides by 4, getting

    dy/(y - 1/2) = 4dt

    Then continue from there.
     
  4. Jan 13, 2009 #3
    Ok, so it should be: ln|y-1/2| = 4t+C

    then: y-1/2 = Ce^4t

    I am not really sure what to do here.
     
  5. Jan 13, 2009 #4
    Do I just set y=3 and t=0 to get C = 2.5?
     
  6. Jan 13, 2009 #5
    You seem to be using C in 2 places which is fine as long as you recognize that they are different things.

    ln|y-1/2| = 4t+C => y - 1/2 = Ae^(4t) where A = e^C
     
  7. Jan 13, 2009 #6
    so,
    y-1/2=e^C*e^(4t)

    Then C still equals 2.5 right?
     
  8. Jan 13, 2009 #7
    Well if t = 0 and y = 3 then 2.5 = e^C*e^(0) => e^C = 2.5 i.e. C = ln(2.5)... how are you getting C = 2.5?

    The point IS C = ln(2.5) is just some constant once again, so I can call that constant. If I defined A = e^C = e^(ln(2.5)) = 2.5 then I would get what you started with. You need to understand that you can't just say C = e^C and say that that equals 2.5, realize which constant you are talking about.
     
  9. Jan 14, 2009 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If y- 1/2= Ce4t and you know that y(0)= 3, then 3-1/2= Ce0= C.
    Yes, C= 5/2= 2.5.

    y(t)= 1/2+ 2.5 e4t which can also be written
    [tex]y(t)= \frac{1+ 5e^{4t}}{2}[/tex]
     
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