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Unformatted text preview: Solutions to the First Midterm Exam Problem 1. a) Differentiating, we get y ′ = x − 2 + x − 1 and y ′′ = 2 x − 3 x − 2 . Substituting that into the equation, we obtain x 2 y ′′ + 2 xy ′ xy = x 2 ( 2 x − 3 x − 2 ) + 2 x ( x − 2 + x − 1 ) x ( x − 1 + ln x ) =2 x − 1 1 2 x − 1 + 2 1 x ln x = x ln x, as required. b) The general solution of the equation is y = integraldisplay ( sin x cos 2 x + x − 2 ) dx = integraldisplay sin x cos 2 x dx + integraldisplay x − 2 dx = cos 3 x 3 x − 1 + C. We have y (2 π ) = cos 3 2 π 3 1 2 π + C = 1 3 1 2 π + C, from which C = 4 3 + 1 2 π . Answer: y = cos 3 x 3 1 x + 4 3 + 1 2 π . c) We observe that √ 1 x 2 is defined for 1 ≤ x ≤ 1. The slope depends on the x coordinate only, always nonnegative, grows from 0 at x = 1 to 1 at x = 0 and then decays from 1 at x = 0 to 0 at x = 1. Hence the solutions will be increasing, concave up for 1 ≤ x ≤ 0 and then concave down for 0 ≤ x ≤ 1. y(0)=1 x y 1 1 1 y(0)=0 1 Problem 2. a) Separating the variables, we get dy 1 2 y = xe 2 x dx = ⇒ integraldisplay dy 1 2 y = integraldisplay xe 2 x dx. Integrating, integraldisplay dy 1 2 y = 1 2 ln  1 2 y  + C and integraldisplay xe 2 x dx = 1 2 integraldisplay x de 2 x = xe 2 x 2 1 2 integraldisplay e 2 x dx = xe 2 x 2 e 2 x 4 + C Hence we get the equation 1 2 ln  1 2 y  = xe 2 x 2 e 2 x 4 + C, that is ln  1 2 y  = xe 2 x + e 2 x 2 + C....
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This note was uploaded on 01/18/2009 for the course MATH 216 taught by Professor Stenstones? during the Spring '07 term at University of Michigan.
 Spring '07
 Stenstones?

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