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Basic Differential Equations Problem (I Have No Idea What I Am Doing)

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Denote by L(t) the length of a fish at time t, and assume that the fish grows according to:
    dL/dt = k(34-L(t)) with L(0) = 2

    a) solve the above equation
    b)Use your solution in part a to determine k under the assumption that L(4) = 10.
    c) Find the length of the fish when t = 10
    d)Find the asymptotic length of the fish; that is, find lim[itex]_{t-->∞}[/itex]


    2. Relevant equations
    I attempted to solve part a by changing the equation to (dL/k(34-L(t))) = dt. I then took the integral of both sides raised the resulting equation to e. My answer was e^(1/-34k)*(ln(Lx) - ln(L(0)). This does not seem right to me but I tried to continue with the rest of the problem.

    For part b, I attempted to solve for k and plug in 10 for L(x) and 2 for L(0). I ended up with k= e^-34*(10-2) = 1.37x10^-14. Once again I do not think this is correct.

    As I mentioned previously, I do not believe I have done any of this problem correctly. I am only in my second quarter of calculus and am struggling. We were introduced to differential equations just the other day and I really do not get how to solve them. Any help would bee greatly appreciated as I have no idea how to proceed. Thank You.
     
  2. jcsd
  3. May 13, 2012 #2

    Office_Shredder

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    When you take the integral of both sides, you have an L(0) which I think is a bit premature. Typically when doing separation of variables it's easier to just do an indefinite integral of both sides (since you don't really know what the bounds of your definite integral are supposed to be). Also your answer involves Ls but no t's, and since you should be solving for L as a function of t what you have written is not a legitimate answer for part a.

    When I do separation of variables I find it helps to keep the logarithm side as simple as possible. Re-write your equation as

    [tex] \frac{ dL} {L-34} = -k dt [/tex]

    Now when you integrate both sides you get
    [tex] \ln( |L-34| ) =-kt+C[/tex]

    This helps to minimize sign and constant errors when there are lots of constants flying around on the harder to integrate side. Can you solve the problem from here?
     
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