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Unformatted text preview: 12 1 Solutions 12. The equation
equilibrium solutions. has no constant solution. Thus, there are no 13. The equation has two constant solutions: and 14. We substitute
for all forces
linear solution is into
and
. . Equality
and the only to get
. Thus and 15. We substitute
into
to get
.
Equality for all means that
must be a constant function,
which can occur only if the coeﬃcient of is 0. This forces
leaving
us with the equation
. This implies
, where is an
integer. Hence
,
is a family of linear solutions. Section 1.3
1. separable; and 2. In standard form we get
. . This is separable; and . We cannot write
3. First write in standard form:
product of a function of and a function of . It is not separable. as a 4. In standard form we get
. We cannot write
product of a function of and a function of . It is not separable. as a 5. Write in standard form to get:
. It is separable;
6. We can factor to get
and . Here we can write
and
.
. It is separable; . 7. In standard form we get
. We cannot write
product of a function of and a function of . It is not separable
8. It is not separable as
of and a function of .
9. In standard form we get: as a cannot be written as a product of a function
It is separable; and 10. The variables are already separated, so integrate both sides of the equation to get
, which we can rewrite as
where
is a constant. Since
, substitute
and
to get that
. Thus the solution is given implicitly by ...
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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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