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Thread: Differential Operators Problem doesn't seem to make sense

  1. #1
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    Differential Operators Problem doesn't seem to make sense

    I have the following problem in an assignment that seems to have some issues.
    The problem is:

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

    So I got the general solution of the ODE by saying $ [(D + 2)(D + 3)]y = 0 $, my result is $ y = A * e^{-2t} + B * e^{-3t} $. Next I had to solve the IVP, so I first found y':
    $ y' = -2tAe^{-2t} -3tBe^{-3t} $.
    Now for y(0) = 2: $2 = A * 1 + B * 1$ or $ A + B = 2 $.
    And for y'(0) = $ \beta $: $ \beta = -2A -3B $.
    I solved for A and B by system of equations and I got: $A = \beta + 6, B = -(\beta + 4) $
    So $y = (\beta + 6)e^{-2t} - (\beta + 4)e^{-3t} $
    And $y' = -2(\beta + 6)e^{-2t} + 3(\beta - 4)e^{-3t}$.

    I solved part a) by setting y' = 0 and solving for t, then plugging t into y. Both were fairly complicated functions of beta but that's not a huge problem.
    EDIT: I hadn't simplified my result for $ y_m $. Now that I did, I find that $ y_m $ simplifies to 0. Since this is the only critical point, does this mean that I must test the boundary points of t to find $t_m, y_m$? If so, assuming $ t \geq 0$, $(t_m,y_m) = (\infty,2)$! Any way I look at this, I can't see how question b) is possible.
    Part b) is where it gets weird. I don't understand how any possible beta can assure that y is greater than 4 for all values of x. As t approaches infinity, $ e^{-2t}, e^{-3t} = 0$, therefore y tends to 0 as t approaches infinity, regardless of the value of beta. Is that not true?

    I haven't tried part c) yet, I'm mostly worried about part b). Have I done something wrong?
    Last edited by wadawalnut; Oct 10th 2015 at 11:51 AM.
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  2. #2
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    Re: Differential Operators Problem doesn't seem to make sense

    a) you have $y^\prime$ as a function of $t$ and $\beta$. Hold $\beta$ fixed and differentiate w/respect to t. Set this equal to zero and solve for t.

    That's the $t_m$, they want. Evaluate $y_m = y(t_m)$ and you are done.

    b) pretty sure they want $y_m > 4$

    c) this is pretty straightforward once you get (b) squared away.
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  3. #3
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    Re: Differential Operators Problem doesn't seem to make sense

    a) Ok thanks, I actually found a mistake in my work and then did the exact procedure you mentioned. However, the only result that produced was the local minimum, not maximum. I got  (t_m, y_m) = (\infty, \frac{\beta}{11} + 2) . Is it possible for tm to be infinity?
    b) At least I'm not crazy. I sent an email to the TA to hopefully get that cleared up.

    Thanks for the reply.
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  4. #4
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    Re: Differential Operators Problem doesn't seem to make sense

    Quote Originally Posted by wadawalnut View Post
    a) Ok thanks, I actually found a mistake in my work and then did the exact procedure you mentioned. However, the only result that produced was the local minimum, not maximum. I got  (t_m, y_m) = (\infty, \frac{\beta}{11} + 2) . Is it possible for tm to be infinity?
    b) At least I'm not crazy. I sent an email to the TA to hopefully get that cleared up.

    Thanks for the reply.
    $\displaystyle{\lim_{\beta->\infty}}t_m = \ln\left(\dfrac 3 2 \right)$

    $\displaystyle{\lim_{\beta->\infty}}y_m =\infty $
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  5. #5
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    Re: Differential Operators Problem doesn't seem to make sense

    Thanks, but that's the answer to c). I was talking about a).
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  6. #6
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    Re: Differential Operators Problem doesn't seem to make sense

    Quote Originally Posted by wadawalnut View Post
    Thanks, but that's the answer to c). I was talking about a).
    I see.

    $t_m(b) = \log \left(\dfrac{3 (b+4)}{2 (b+6)}\right)$

    after a bunch of algebra you can obtain

    $y_m(b) = \dfrac{4 (b+6)^3}{27 (b+4)^2}$
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