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epsilonzero
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Homework Statement
Consider [itex] x = Ae^{at} + Be^{bt} [/itex]
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
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] [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.
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