Sunday, December 30, 2012

Having trouble understanding how to get started on the attached problem in Asymptotics. Any help is greatly appreciated.

You need to write the characteristic equation, such that:


`r^2 - (1/2)r + 1/16+x-e^x = 0`


You need to evaluate `Delta` , such that:


`Delta = (-1/2)^2 - 4(1/16+x-e^x) = 1/4 - 1/4 - 4x + 4e^x`


`Delta = 4e^x - 4x`


The characteristic equation has real solutions if `Delta > 0`


`4e^x - 4x > 0 => e^x - x > 0 => e^x > x`


Attaching the equation yields:


`e^x = x => ln e^x = ln x => x = ln x => (ln x)/x = 1`


If `x -> oo` yields `lim _(x->oo) (ln x)/x = 0`


Hence,` r_(1,2) = (1/2 +-sqrt(4(e^x - x)))/2 => r_(1,2) = (1/2 +-2sqrt(e^x - x))/2 => r_(1,2) = 1/4 +- sqrt(e^x - x) `


If `x->oo` yields `r_(1,2) = 1/4 +- oo => r_(1,2) = +-oo`


Hence, the general solution to the equation, for `x->oo` , is:


`y(t) = c_1*e^(+oo*t) + c_2*e^(-oo*t)`

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