I first find its general solution

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khan (fak337) – HW - 9.1, 9.2, 9.3 – nibert – (52800) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 5.0points If y 1 satisfies the equations y + 4 y 2 ln x = 0 , y (1) = 1 6 for x > 1 2 , determine the value of y 1 ( e ). 1. y 1 ( e ) = 1 11 2. y 1 ( e ) = 1 9 3. y 1 ( e ) = 1 10 correct 4. y 1 ( e ) = 1 13 5. y 1 ( e ) = 1 12 Explanation: After separating variables and integrating, the given differential equation becomes integraldisplay 1 y 2 dy = 1 y = integraldisplay 4 ln x dx. To integrate the right hand side we integrate by parts. Thus 1 y = 4( x ln x x ) + C with C an arbitrary constant. For y 1 the value of C is determined by the condition y (1) = 1 6 since y (1) = 1 6 = C = 10 . Consequently, y 1 ( x ) = 1 4( x ln x x ) + 10 . At x = e , therefore, y 1 ( e ) = 1 10 . 002 5.0points Solve dy dx = 8 x 15 y 1 + x 16 for y. 1. arcsin(1 + x 16 ) + C 2. arctan(1 + x 16 ) + C 3. None of these correct 4. arccos(1 + x 16 ) + C Explanation: y = 8 x 15 y 1 + x 16 y y = 8 x 15 1 + x 16 integraldisplay y y dx = integraldisplay 8 x 15 dx 1 + x 16 ln | y | = 1 2 integraldisplay 16 x 15 dx 1 + x 16 = 1 2 ln | 1 + x 16 | + C 0 y = e 1 2 ln | 1+ x 16 | + C 0 = e ln 1+ x 16 e C 0 y = C radicalbig 1 + x 16 003 5.0points Use the direction field of the differential equation y = y 3 to sketch a solution curve that passes through the point (0 , 1).
khan (fak337) – HW - 9.1, 9.2, 9.3 – nibert – (52800) 2 1. 2 2 2 2 x y 2. 2 2 2 2 x y 3. 2 2 2 2 x y 4. 2 2 2 2 x y correct 5. 2 2 2 2 x y 6. 2 2 2 2 x y Explanation: The vector field for y = y 3 with a sample curve through our initial condition y (0) = 1 looks like the following 2 2 2 2 x y 004(part1of3)5.0points For the differential equation dy dx = 6 x 2 y,
khan (fak337) – HW - 9.1, 9.2, 9.3 – nibert – (52800) 3 (i) first find its general solution. 007 5.0points
khan (fak337) – HW - 9.1, 9.2, 9.3 – nibert – (52800) 4 If y 0 is the particular solution of dy dx = 4 x 3 y, y (0) = 4 , find the value of y 0 (2). 1. y 0 (2) = 4 e 16 correct 2. y 0 (2) = 4 e 16 3. y 0 (2) = 4 e 8 4. y 0 (2) = 4 e 5. y 0 (2) = 4 e 8 Explanation: The differential equation becomes integraldisplay dy y = integraldisplay 4 x 3 dx after separating variables and integrating. Thus its general solution is ln y = x 4 + C, which can also be written as y = e C e x 4 = Ae x 4 with A an arbitrary constant. The value of A

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