Find Initial value such that solutions go to -infinity

1. Mar 18, 2015

k_squared

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. Mar 18, 2015

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.

3. Mar 18, 2015

k_squared

Yes, sorry. That is what I meant. I thought it was logically unambiguous, but yes Thank you, that is what I meant..

4. Mar 18, 2015

SammyS

Staff Emeritus
You might consider writing that out correctly.

5. Mar 18, 2015

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.

6. Mar 19, 2015

k_squared

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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