Unformatted text preview: 1 Solutions 13. The given equation is in standard form,
, and the integrating factor is
by the integrating factor to get 21 , an antiderivative is
. Now multiply 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. To satisfy the initial condition,
, so
. Thus, the solution of the initial value problem is
14. The given equation is in standard form,
, 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
is a constant. Now multiply by
to get
the general solution. where
for 15. The given linear diﬀerential equation is in standard form,
,
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 and taking antiderivatives of both sides gives
is a constant. Now multiply by to and we get
for the general solution. Letting
gives
and . Thus where
so ...
View
Full
Document
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

Click to edit the document details