Find Initial value such that solutions go to -infinity

[k_squared]

Homework Statement

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.

Homework Equations

(All are given in the problem statement.)

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.

 

[Mark44]

Homework Statement

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.

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.

Homework Equations

(All are given in the problem statement.)

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.

 

[k_squared]

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

 

[SammyS]

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

You might consider writing that out correctly.

 

[Mark44]

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

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 -4e^t, which grows negatively as t → ∞. With this help, maybe you can figure out what happens when y₀ < -16/3 and when y₀ > -16/3.

 

[k_squared]

If y⁰ = -16/3, then C = 0, so the term with $e^{3t/2}$ drops out, and the dominant term is -4e^t, which grows negatively as t → ∞. With this help, maybe you can figure out what happens when y₀ < -16/3 and when y₀ > -16/3.

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!