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)`
No comments:
Post a Comment