Unformatted text preview: 30 1 Solutions 2. Since the numerator and denominator are homogeneous of degree 1 the
quotient is homogeneous. Let
. Clearly, . Then and so is an equilibrium solution. Separating the variables and using partial fractions gives . Integrat ing gives
. Simplifying and exponentiating
gives
,
. Now let
then
for
. The equilibrium solution
gives
as another solution.
3. Since the numerator and denominator are homogeneous of degree 2 the
quotient is homogeneous. Let
. Then
. So
. There are two equilibrium solutions
. Separating the variables and using partial fractions gives
. Integrating and simplifying gives
. Solving for gives for
When
equilibrium solution , for , and so , we get
or
, which is the same as the
. The equilibrium solution
gives
. Thus we can write the solutions as , and . 4. Since the numerator and denominator are homogeneous of degree 2 the
quotient is homogeneous. Let
. Subtract . Then from both sides to get . Separate the variables to get Integrating gives
, and simplify to get , . Now exponentiate, substitute
. 5. Since the numerator and denominator are homogeneous of degree 2 the
quotient is homogeneous. Let
from both sides to get
. Separating variables gives . Then . Subtract . The equilibrium solutions are
and integrating gives . Exponentiating gives
we get
. Now
so
solutions
become
. These occur when
included in the general formula. and by simplifying
. The equilibrium
, so are already ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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