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exam1-sol(1)

# exam1-sol(1) - Exam 1 MAP 2302 Name Student ID This is a...

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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 = ± e 1 2 x 2 2 . These are two functions y ( x ), while we know there should only be one (by the existence and uniqueness theorem). So which one is the solution to our problem? That’s easy: the one

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exam1-sol(1) - Exam 1 MAP 2302 Name Student ID This is a...

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