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Unformatted text preview: at . Rearranging, we have x + C K xe at = Ce at , or x ´ K + Ce at K µ = Ce at . Solving for x we arrive at the explicit form x = KCe at K + Ce at = 1 1 C e at + 1 K . We can now substitute for C and simplify to get x ( t ) = 1 ± Kx o Kx o ² eat + 1 K = Kx o ( Kx o ) eat + x o . NOTES: From this explicit form, we can see that if x o = 0 then x ( t ) ≡ 0r, while if x o > then lim t →∞ Kx o ( Kx o ) eat + x o = Kx o x o = K . So we see that the asymptotic behavior predicted from the analysis of the direction ﬁeld is also evident from the form of the solution....
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This note was uploaded on 12/02/2009 for the course MATH 352 taught by Professor Staff during the Spring '08 term at University of Delaware.
 Spring '08
 Staff

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