# Constant coefficient equation

1. Sep 22, 2008

### epsilonzero

1. The problem statement, all variables and given/known data

Consider $x = Ae^{at} + Be^{bt}$

a) Show this can have at most one zero.
b) Show that if x(t)=0 for t>0, then either x(0)>0 and x'(0)<0; or x(0)<0 and x'(0)>0

3. The attempt at a solution

a)

x = Ae^{at} + Be^{bt}
0 = Ae^{at} + Be^{bt}
0 = ln(Ae^{at} + Be^{bt}}
0 = ln(A) + ln(e^{at}) + ln(B) + e^{bt}
0 = ln(A) + ln(B) + at + bt
0 = ln(AB) + t(a+b)
t=\frac{-ln(AB)}{a+b}

A,B,a, and b are constants so that's a distinct solution and the equation has just one zero, right?

b)

t>0
a,b<0 (I think that was given in class)

x = Ae^{at} + Be^{bt} [/itex] x' = aAe^{at} + bBe^{bt} so after some manipulation I got x = ln(AB) + t(a+b) x' = ln(abAB) + t(a+b) In both these equations, I think the first term would be positive and the second term negative. I don't see how to reach the "either x(0)>0 and x'(0)<0; or x(0)<0 and x'(0)>0" conclusion. Any help? Thank you very much. p.s. I was trying to use latex for this post but couldn't get line breaks to work. How do I get equations on different lines? I tried \\ but nothing happened. Last edited: Sep 22, 2008 2. Sep 22, 2008 ### statdad For your "equations on lines question" - start with the usual tex tags at beginning and end - you need to use \begin{align*} and \end{align*} inside the tex tags. -Put one equation on each line: use an ampersand ("&") at the spot where you wish the equations to line up. On this forum I typically put this before an equal or inequality sign - To end one line and begin a new one, use // at the end. Do not do this at the end of your final equation, or you will have extra blank space When it comes to your work there is an error in the second line: note that $$\ln{(Ae^{at} + Be^{bt})} \ne \ln A + \ln{e^{at}} + \ln B + \ln{e^{bt}}$$ even though your work states the two sides are equal. It is impossible to simplify $$\ln(X + Y)$$ into sums. (I foolishly hit Post rather than preview, so I have to edit my message to finish) so that all of your work after this point is incorrect. To address the question about the zero: perhaps start this way (perhaps in the line of a suggestion to you rather than an open question) (by the way, I'm using the align* idea in the work below) Begin by assuming there is a solution. \begin{align*} Ae^{at} + Be^{bt} & = 0\\ e^{at} \left(A + Be^{b-a}t\right) & = 0\\ \end{align*} The first factor above can't be zero, so the second one, in parentheses, must be. That will give you some information about the unknown constants in your question. Last edited: Sep 22, 2008 3. Sep 22, 2008 ### epsilonzero I don't see how [itex] A + Be^{(b-a)t} = 0 would tell me that there is at most one zero. Could you elaborate a little?

Also any ideas on part b?

Thanks

4. Sep 22, 2008

### HallsofIvy

Staff Emeritus
Did you really try to solve $A+ Be^{(b-a)t}= 0$? What happens? How many roots do you get.

5. Sep 22, 2008

### epsilonzero

I got t=(ln(-A/B))/(b-a)

So I just say that's at most 1 root if either A or B is negative?

For the second part I have

x(0) = Ae^(a*0)+Be^(b*0) = A + B
x'(0) = Aae^(a*0)+Bbe^(b*0) = Aa + Bb

I know 0=Ae^(a*t)+Be^(b*t) so Ae^(at)=-Be^(bt)

a,b<0 and t>0

Now I'm stuck at what to do next. I don't know anything about the relative sizes of a, b, A and B so I can't make any conclusion about the signs of x(0) and x'(0). Any ideas?

Last edited: Sep 22, 2008