# Find Initial value such that solutions go to -infinity

• k_squared
In summary, when y0=-16/3, the equation becomes y=-4e^t+16/3-2t+Ce^(-3t/2), and when y0>-16/3, the equation becomes y=-4e^t+16/3+2t+Ce^(3t/2).
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.

k_squared said:

## 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.
k_squared said:

## 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.

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

k_squared said:
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.

k_squared said:
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 -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.

Mark44 said:
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.

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!

## 1. What does it mean for a solution to go to -infinity?

When a solution goes to -infinity, it means that the value of the solution is becoming infinitely negative as it approaches infinity. In other words, the solution is decreasing without bound.

## 2. Why is it important to find the initial value such that solutions go to -infinity?

It is important to find the initial value because it helps determine the behavior of the solution as it approaches infinity. If the initial value is not set correctly, the solution may not go to -infinity as desired.

## 3. How do you find the initial value such that solutions go to -infinity?

To find the initial value, you can use the limit definition of a function. This involves setting the limit as x approaches infinity to -infinity and solving for the initial value.

## 4. Can the initial value be any number if we want the solutions to go to -infinity?

No, the initial value must be chosen carefully in order for the solutions to go to -infinity. If the initial value is too large, the solution may not go to -infinity. It must be chosen based on the specific function and its behavior.

## 5. Are there any real-world applications of finding initial value such that solutions go to -infinity?

Yes, this concept is often used in mathematical modeling and optimization problems. For example, in economics, finding the initial value that leads to solutions going to -infinity can help determine the optimal production level for a company that wants to minimize costs.

• Calculus and Beyond Homework Help
Replies
2
Views
559
• Calculus and Beyond Homework Help
Replies
2
Views
500
• Calculus and Beyond Homework Help
Replies
3
Views
568
• Calculus and Beyond Homework Help
Replies
11
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
825
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
11
Views
1K