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Unformatted text preview: 18 1 Solutions for the general solution of the equation. To ﬁnd the constant to satisfy
the initial condition
, substitute
into the general solution
to get
. Hence
, and the solution of the initial
value problem is 2. Divide by to put the equation in the standard form In this case
, an antiderivative is
, and the
integrating factor is
. (We do not need
since we are
working near
where
.) Now multiply by the integrating
factor to get
, the left hand side of which
is a perfect derivative. Thus
and taking antiderivatives
of both sides gives
where
is a constant. Now
multiply by
to get
for the general solution.
Letting
gives
so
and 3. This equation is already in standard form. In this case
antiderivative is
, and the integrating factor is
Now multiply by the integrating factor to get the left hand side of which is a perfect derivative
and taking antiderivatives of both sides gives where
is a constant. Now multiply by
the general solution. Letting
gives 4. Divide by , an
. . Thus to get
so for to put the equation in the standard form In this case
, an antiderivative is
, and the integrating
factor is
. Now multiply the standard form equation by the integrating factor to get
, the left hand side of which is a perfect ...
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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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