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Ordinary Diff Eq Exam Review Solutions 113

Ordinary Diff Eq Exam Review Solutions 113 - 1 Solutions...

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1 Solutions 115 1 2 3 1 1 2 3 4 5 6 7 t y 3. This function is g ( t 1) h ( t 1) where g ( t ) = t , so the graph of f ( t ) is the graph of g ( t ) = t translated 1 unit to the right and then truncated at t = 1 , with the graph before t = 1 replaced by the line y = 0 . Thus the graph is 0 1 2 0 1 2 3 t y 4. This function is g ( t 2) h ( t 2) where g ( t ) = t 2 , so the graph of f ( t ) is the graph of g ( t ) = t 2 translated 1 unit to the right and then truncated at t = 2 , with the graph before t = 2 replaced by the line y = 0
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