Unformatted text preview: 32 1 Solutions and solve for . The initial condition implies to get and so . The solution is thus . 10. Note that
is a solution. Divide by
to get
. Let
. Then
and substituting gives
. In the
standard form for linear equations this becomes
. Multiplying
by the integrating factor
gives
so that
.
Hence
. Now go back to the original function by solving
for . Thus
. The general solution
is
and
.
11. Note that is a solution. First divide both sides by . Let . Then , so . Substituting , which in standard form is gives to get . An integrating factor is
so that
. Integrating both
sides gives
, where the integral of the right hand side is
done by the substitution
. Solving for gives
. Since
we ﬁnd
12. Note that .
is a solution. First divide both sides by
. Let . Then stituting gives , so to get
. Sub , which in standard form is . An integrating factor is
Thus
Integrating both sides gives
, where the integral of the
right hand side is computed using integration by parts with
and
. Solving for gives
. Since
we
ﬁnd . 13. Note that is a solution. Divide by
. Let gives . Then and to get
and substituting . In standard form we get
. Multiplying by the integrating factor ...
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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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