Unformatted text preview: 114 1 Solutions . The initial conditions
. By continuity it follows that
and these constitute the initial
values for the equation
on the interval
general solution on this interval is
Solving for and gives
Putting the two pieces together gives
and if 33. 1.
2. It is enough to observe that
does not exist. But letting
for all positive integers , while letting
so there is one sequence
cannot be continuous at 0.
3. To be piecewise continuous,
would have to have a limit at approaches 0 from above, and this is not true as shown in part 2. Section 5.2
if 1. Thus, the graph is if
graph is Thus, the ...
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- Fall '08
- Topology, Continuous function, initial conditions, positive integers, general solution