Ordinary Diff Eq Exam Review Solutions 112

# Ordinary Diff Eq Exam Review Solutions 112 - 114 1...

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Unformatted text preview: 114 1 Solutions . The initial conditions and imply that . So on . By continuity it follows that and and these constitute the initial values for the equation on the interval . The general solution on this interval is and at we get and . Solving for and gives and so that . Putting the two pieces together gives and if 33. 1. for all , while . 2. It is enough to observe that does not exist. But letting gives for all positive integers , while letting gives so there is one sequence with while another sequence with so 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 if if 1. Thus, the graph is if if if if 2. graph is Thus, the ...
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