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Unformatted text preview: 1 Solutions 19 derivative
. (Note that this is just the original left hand side of the
equation. Thus if we had recognized that the left hand side was already a
perfect derivative, the preliminary steps could have been skipped for this
problem, and we could have proceeded directly to the next step.) Thus
the equation can be written as
and taking antiderivatives of
both sides gives
where
is a constant. Now divide by to
get for the general solution.
. Now let 5. The general solution from Problem 4 is
. So
and
.
6. Divide by to get to put the equation in the standard form , an antiderivative is
, and the
In this case
integrating factor is
. Now multiply the standard form equation
by the integrating factor to get
, the left hand
side of which is a perfect derivative
. Thus
To
integrate
we consider the cases
and
separately.
Case A simple substitution gives : Hence, . and so Case Use integration by parts to get :
. Then . 7. We ﬁrst put the equation in standard form and get In this case
factor is , an antiderivative is
, and the integrating
. 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 divide by
general solution. . Thus to get for the
new: please check 8. In this case
multiply to get and the integrating factor is
, which simpliﬁes to . Now ...
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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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