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Differential Linear Operator Problem not making sense

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    I think there may be something wrong with a problem I'm doing for homework. The problem is:

    Solve the IVP with the differential operator method:
    [itex] [D^2 + 5D + 6D], y(0) = 2, y'(0) = \beta > 0[/itex]
    a) Determine the coordinates [itex](t_m,y_m)[/itex] of the maximum point of the solution as a function of [itex]\beta[/itex].
    b) Determine the smallest value of [itex]\beta[/itex] for which [itex]y \geq 4 [/itex].
    c) Determine the behavior of [itex] t_m [/itex] and [itex]y_m[/itex] as [itex]\beta \rightarrow \infty [/itex].


    2. Relevant equations

    3. The attempt at a solution
    So I solved the differential equation by simplifying the operator to [itex] [D^2 + 11D] [/itex] (I thought it was weird that I had to do such a simple simplification). Then I wrote [itex] (D + 11)(D)y = 0 [/itex]. Next I said let [itex] z = [D]y [/itex] so the problem simplifies to [itex] [D + 11]z = 0 [/itex]. From this I got that [itex] z = c_1e^{-11t} [/itex] from separation of variables. Then, since z is just y', I integrated z to find y, which gave me [itex] y = \dfrac{-c_1}{11}e^{-11t} + c_2 [/itex]. According to Wolfram Alpha, so far, so good... (so what?).

    Next, I solved the initial value problem.
    [itex] y'(0) = \beta [/itex] so [itex]\beta = c_1e^0 [/itex] and [itex] c_1 = \beta [/itex].
    [itex] y(0) = 2 [/itex] so [itex] 2 = \dfrac{-\beta}{11} \beta e^0 + c_2[/itex] and [itex] c_2 = \dfrac{\beta}{11} + 2[/itex].

    Now I have:
    [itex] y = \dfrac{-\beta}{11} e^{-11t} + \dfrac{\beta}{11} + 2 [/itex]
    [itex] y' = \beta e^{-11t} [/itex]

    Now for part a).
    To find the maximum, I set y' to 0 and solved for t. I got:
    [itex] 0 = \beta e^{-11t} [/itex]. This is not solveable. I'm assuming [itex] t \geq 0 [/itex] because conventionally t is time and must be greater than 0. Since beta is positive, the maximum of y occurs when [itex] t = \infty [/itex], so [itex] (t_m,y_m) = (\infty, \dfrac{\beta}{11} + 2)[/itex]. This seems sketchy but is the most reasonable answer I can come up with.

    And now the very strange part, part b).
    How is it possible for me to find a value for beta such that [itex] y \geq 4[/itex] if THE PROBLEM STATES THAT [itex]y(0) = 2[/itex]?
    Isn't that a contradiction? I started this problem assuming that condition, and now my function of y prohibits the constraint that [itex]y \geq 4 [/itex]. Am I missing something?

    I haven't really tried part c) because I'm assuming I already messed everything up, I just can't figure out where.
    Any help is appreciated.
     
  2. jcsd
  3. Oct 10, 2015 #2

    Mark44

    Staff: Mentor

    Typos above? I think it should be ##[D^2 + 5D + 6]y = 0##, and then the initial conditions.
    See my comment above. I'm almost certain that the DE is y'' + 5y' + 6y = 0. If so, it can be solved fairly easily.
     
  4. Oct 10, 2015 #3
    I did not make a typo, that was the actual question.
     
  5. Oct 10, 2015 #4

    Mark44

    Staff: Mentor

    I didn't say or even imply that you made the typo, but I am almost certain (99.44%) it is a typo.
     
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