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Unformatted text preview: Exam 1: MAP 2302 * September 26, 2008 Name : Student ID : This is a closed book exam and the use of calculators is not allowed. 1. Solve the IVP: dy dx = xy, y (1) = 1 . By the existence and uniqueness theorem, we know there is a unique solution. Separable equation: 1 y dy = xdx, ln  y  = x 2 2 + C, C arbitrary constant ,  y  = C e x 2 2 , C positive constant (since C = e C ) , y = C e x 2 2 = C e x 2 2 , C nonzero constant Since y (1) = 1, it follows that 1 = C e 1 2 or C = e 1 2 , and thus y ( x ) = e 1 2 x 2 2 Common mistake : After finding  y  = C e x 2 2 , one determines C , using the initial condition:   1  = C e 1 2 , so that C = e 1 2 . So far, so good, but then one drops the absolute value and concludes: y = e 1 2 x 2 2 . This is wrong because y (1) = +1 instead of 1. The mistake made is that the absolute value cannot be dropped. A careful argument is as follows: y =...
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This note was uploaded on 09/12/2011 for the course MAP 4305 taught by Professor Deleenheer during the Summer '06 term at University of Florida.
 Summer '06
 DeLeenheer

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