midterm1sol - Solutions to the First Midterm Exam Problem...

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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 ) =- cos 3 2 3- 1 2 + C =- 1 3- 1 2 + C, from which C = 4 3 + 1 2 . Answer: y =- cos 3 x 3- 1 x + 4 3 + 1 2 . c) We observe that 1- x 2 is defined for- 1 x 1. The slope depends on the x coordinate only, always non-negative, 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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midterm1sol - Solutions to the First Midterm Exam Problem...

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