midterm1sol

# midterm1sol - Solutions to the First Midterm Exam Problem 1...

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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 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 1...

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