Unformatted text preview: 14 1 Solutions 17. There are two equilibrium solutions;
ables and using partial fractions gives
and simplifying gives and . Separating vari. Integrating which is equivalent to , a nonzero constant. Solving for gives
. When
we get
the equilibrium solution
. However, there is no which gives the
other equilibrium solution
.
18. There are no equilibrium solutions. Separating variables gives
and integrating gives
. Solving for
gives
, where
.
19. Separating variables gives
and integrating gives
Thus
, a real constant.
20. Separating variables gives
. Simplifying gives . and integrating gives
, a real constant. 21. In standard form we get
from which we see that
is an equilibrium solution. Separating variables and simplifying gives
. Integrating and simplifying gives
.
22. Separating variables gives
a constant. and integrating gives , 23. The equilibrium solution is
. Separating variables gives
. Integrating and simplifying gives
, real constant.
24. In standard form we get
from which we see
and
are equilibrium solutions. The equilibrium solution
satisﬁes the
initial condition
so
is the required solution.
25. is the only equilibrium solution. The equilibrium solution
satisﬁes the initial condition
so
is the required solution. 26. Rewriting we get
from which we see that
is
an equilibrium solution. Separating variables gives
and
integrating gives
, a constant. Solving for by
taking the exponential of both sides gives
, and allowing
gives the equilibrium solution. The initial condition gives
so
. Thus
.
27. In standard form we get
ables and integrating gives so is a solution. Separating vari. Solving for gives ...
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 Fall '08
 BELL,D
 Algebra, Equilibrium, Exponential Function, Equilibrium point, equilibrium solution

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