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Unformatted text preview: 1 Solutions so that
for
Putting these three pieces together, we ﬁnd that the solution is 113 . if
if
if
if
28. The general solution of
on any interval is found by using the
integrating factor . Using this integrating factor the general solution of
on the interval
is found to be
. The initial condition actually occurs at the end of this interval, but
by continuity we can substitute
in this formula to get
so
. Hence on the interval
the solution is
. The
general solution of
on the interval
is
and
this gives
so
and
on
. Putting these two pieces together, we ﬁnd that the solution is
if
if
29. The characteristic polynomial of the equation
is
so the homogeneous equation has the solution
for constants and . On the interval
the equation
has a particular solution
so the general solution has the form
. The initial conditions give
and
. Solving gives
,
so
on
. By continuity it follows that
and
and these constitute the initial values for the equation
on the interval
. The general solution on this interval is
and at
we get
and
. Solving for and gives
and
so that
. Putting
the two pieces together gives
if 30. The characteristic polynomial of the equation
is
so the homogeneous equation has the solution
for constants and , which is valid on the interval ...
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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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