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Unformatted text preview: 112 1 Solutions . The general solution of
on
is
.
Thus
and solve for to get
,
so that
for
. Continuity will then give
, which will provide the initial condition for
the next interval
. The general solution to
on
is
and the constant is obtained from the initial condition
, which gives
, so
that
for
. Putting these three pieces
together, we ﬁnd that the solution is
if
if
if 26. The general solution of
on the interval
is found by using
the integrating factor
. The general solution is
and the initial condition gives
. By continuity we must have
, which will provide the initial condition for
the next interval
. The general solution of
on
is
. Setting
found from
the interval
equal to
found from the interval
and solving for gives
. Thus the complete solution is
if
if 27. The general solution of
on any interval is found by using
the integrating factor
. The general solution on the interval
is
and since the initial condition is
, the solution on
is
. Continuity then given
, which will be the initial
condition for the interval
. The general solution of
on the interval
is
and the initial condition
gives
so that
. Thus
for
. Continuity of
at
will then give
, which will provide the initial condition for the next interval
.
The general solution of
on
is
. Thus
and solve for to get
, so that
for
.
Continuity will then give
, which will provide the initial
condition for the next interval
. The general solution to
on
is
and the constant is obtained from the initial
condition
, which gives
, ...
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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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