WrittenHW01-Solutions

WrittenHW01-Solutions - Math 216 (Section 50) Written...

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Unformatted text preview: Math 216 (Section 50) Written Homework #1 Solutions Fall 2007 1 ( 1.14 ) y 1 y 1 = 3 e 3 x = y 00 1 = 3 2 e 3 x = 9 e 3 x |{z} y 1 = 9 y 1 . Therefore y 1 is a solution of the differential equation y 00 = 9 y . y 2 y 2 =- 3 e- 3 x = y 00 2 = (- 3) 2 e- 3 x = 9 e- 3 x | {z } y 2 = 9 y 2 . Therefore y 2 is a solution of the differential equation y 00 = 9 y . 2 ( 1.210 ) dy dx = xe- x = dy = xe- x dx = Z dy = Z xe- x dx Now, integration by parts yields Z xe- x dx = x (- e- x )- Z (- e- x ) dx =- xe- x + Z e- x dx =- xe- x- e- x + C =- ( x + 1) e- x + C. Thus Z dy = Z xe- x dx = y ( x ) =- ( x + 1) e- x + C. However, since y (0) = 1, plugging x = 0 into the above equation, y (0) =- 1 + C = 1 = C = 2 . Therefore the solution is given by y ( x ) =- ( x + 1) e- x + 2 . 3 ( 1.316 ) Let us write f ( x,y ) = x- y. Then our differential equation can be written as dy dx = f ( x,y ) ....
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This homework help was uploaded on 04/02/2008 for the course ENGR 101 taught by Professor Ringenberg during the Fall '07 term at University of Michigan.

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WrittenHW01-Solutions - Math 216 (Section 50) Written...

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