Solving Exponential Rates: Intervals & Solutions

[epsilonzero]

Hey, I'm having trouble starting this assignment. If you could just tell me how to get started and the path I should be on then that would be great and I can do the rest. Here are the questions:

1) Solutions to x= - x^k for x0>0 satisfy x(t) goes to 0 as t goes to infinity. Characterize the rate at which these solutions go to 0.

2) How do things change for x'= -m x^k for m>0?


I think question 1 will have several answers for various intervals but I'm how do I find these intervals?

 

[HallsofIvy]

The first problem makes no sense at all. "Solutions to x= -x^k" are numbers that do not "go" anywhere as t goes to infinity. I assume you mean "dx/dt= -x^k"?

In that case you should be able to integrate dx/x^k= -dt without difficulty. The same is true for x'= -mx^k. Integrate dx/x^k= -mdt. The case k= 1 will have to be treated separately.

 

[epsilonzero]

I just copied and pasted so yeah, it seems like there were a few problems. The x should have been an x'. Also I think the question is asking what rate does x(t)->0.

What I've done so far is

x'=-x^k
(-x^-k)dx=-dt
-Int(x^-k)dx=Int(1)dt
1/(-k+1)*x^(-k+1)=t+C

Where do I go from here to find the rate at which x(t) goes to 0?


For number 2 I have

x'=-mx^k
(x^-k)dx=-mdt
Int(x^-k)dx=-Int(m)dt
1/(-k+1)*x^(-k+1)=-mt

Now how do I characterize the effect m has? I see that k=1 will have to be treated separately because there will be 0 in the denominator.

Thank you for your help.

 

[epsilonzero]

I'm still at this step: x=(t(k-1)+C)^(1/(1-k)) and I want to know at what rate x(t) goes to 0 for different intervals of k. I'm not sure how to figure out what these intervals should be or how to calculate the rate.

I think for rate there's an equation (x(t)-a)/(e^mt) as t->infinity where m is the rate and a is an x value. But I don't know what to use for a. Would a be 0?

Then the second equation is x=(mt(k-1)+C)^(1/(1-k)). What effect does m have? Would it just speed up the rate that x(t) approaches 0?

Any help is appreciated.