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Find Initial value such that solutions go to -infinity

  1. Mar 18, 2015 #1
    1. The problem statement, all variables and given/known data
    For the equation y=-4e^t -4/3-2t+Ce^(3t/2), find the initial value y(0)=y_0 that separates solutions that grow positively as t -> infinity from those that grow negatively.

    2. Relevant equations

    (All are given in the problem statement.)

    3. The attempt at a solution
    I have half a large page scribbled in small writing over this. I attempted things like Ce^(3t/2) >= 4e^t+2t, which made a pretty graph on Wolfram-alpha. However, the closest I got was this: since t=0, the only thing that can vary in the equation is C. y_0=-4-0-4/3+C. C=y_o+16/3.

    Sadly, however, this seems to be utterly useless. The answer is -16/3. How would one come to this answer?

    Thanks.
     
  2. jcsd
  3. Mar 18, 2015 #2

    Mark44

    Staff: Mentor

    What you wrote is ##y = -4e^t - \frac 4 3 - 2t + Ce^{3t/2}##. Is that what you meant? If not, use parentheses around the entire numerator and entire denominator.
     
  4. Mar 18, 2015 #3
    Yes, sorry. That is what I meant. I thought it was logically unambiguous, but yes Thank you, that is what I meant..
     
  5. Mar 18, 2015 #4

    SammyS

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    Staff Emeritus
    Science Advisor
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    Gold Member

    You might consider writing that out correctly.
     
  6. Mar 18, 2015 #5

    Mark44

    Staff: Mentor

    Then what you wrote was correctly written. We get so many people here who write things like x - 2/x2 - 4, when what they actually mean is ##\frac{x - 2}{x^2 - 4}##. I thought you had meant ##\frac{-4e^t - 4}{3 - 2t + Ce^{3t/2}}##.

    Anyway, it looks to me like you're on the right track. y(0) = y0 = -4 - 4/3 + C, so solving for C yields C = y0 + 16/3.

    If y0 = -16/3, then C = 0, so the term with ##e^{3t/2}## drops out, and the dominant term is -4et, which grows negatively as t → ∞. With this help, maybe you can figure out what happens when y0 < -16/3 and when y0 > -16/3.
     
  7. Mar 19, 2015 #6
    And I see the world more clearly! When C is *not* zero, the exponential with C becomes the dominant term because of it's greater exponent, which is positive as t goes to infinity. However, the exponential with the -4 is negative as e goes to infinity. Therefore, on must *eliminate* C in order to switch the direction!
     
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