Unformatted text preview: 1 Solutions 31 6. Since
and are homogeneous of degree 2 their quotient
is a homogeneous function. Let
. Then
and the given
diﬀerential equation becomes Simplifying and separating variables gives
get
and so
. for it is easy to see that is homogeneous. Let
. Clearly . Then . are equilibrium solution.
. Integrating gives Separating variables gives
and so . Integrating we
we get . Since 7. In standard form we get Simplifying gives . Since . Now substitute
. The equilibrium solutions imply to get
are also solutions.
8. In standard form we get . It is straightforward to see is homogeneous of degree one. So that is a homogeneous diﬀerential equation. Let
or
. It follows that . Integrating rium solution. Separating variables gives
gives then
is an equilib . (To integrate the left hand side use the trig substitution
. Now let .) Exponentiating gives
. Then , . The case where gives the equilibrium solution.
9. Note that although
is part of the general solution it does not satisfy
the initial value. Divide both sides by
to get
. Let
. Then
. Substituting gives
or
.
An integrating factor is . So
. Integrating both sides gives
, where we have used integration by parts to compute
. Solving for gives
. Now substitute ...
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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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