MATH 2410Q Midterm: Exam 1 Spring 2017 Solutions

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31 Jan 2019
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Clearly mark each answer: (20 points) check if the given function y is a solution of the initial value problem (a) y = x2(1 + ln x) y (x) = 3xy (x) 4y x2 y(e) = 2e2, y (e) = 5e. First we check the initial conditions. y(e) = e2(1 + ln e) = 2e2. Using the product rule we nd y (x) = 2x(1 + ln x) + x and as a result y (e) = 2e(1 + ln e) + e = 5e. Again using the product rule, we compute y (x) = 2 + 2 ln x + 2 + 1 = 5 + 2 ln x. 3x(2x(1 + ln x) + x) 4x2(1 + ln x) x2. That shows that y = x2(1 + ln x) is a solution to. = 6 + 6 ln x + 3 4 4 ln x = 5 + 2 ln x. y (x) =

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