Unformatted text preview: 24 1 Solutions 24. In this case the given diﬀerential equation is in standard form. We have
, an 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
. Thus
and taking antiderivatives of both sides gives
where
is a constant. However, the right hand side has no closed form
antiderivative. Using Corollary 8 to can write 25. Divide by to put the equation in the standard form In this case
, an 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
.
Thus
and taking antiderivatives of both sides gives
for
where
is a constant. Now multiply by
to get
the general solution. Letting
gives
so
and gal
lbs
min
gal
gal
output rate
min 26. input rate: input rate
output rate:
Since input rate lbs
.
min
lbs
gal lbs
.
min output rate we get the initial value problem Simplifying and putting in standard form gives . The coeﬃ cient function is
,
, and the integrating
factor is
. Thus
. Integrating and simplifying gives
The initial condition implies
so
. The
concentration of the brine solution is now obtained by dividing by the
volume which is
gallons:
. ...
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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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