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Fall 2006 - Takeda's Class - Practice Exam 2

# Fall 2006 - Takeda's Class - Practice Exam 2 - Math 20D...

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Math 20D, Sample Midterm 2 Solutions November 11, 2006 Name: Section: This exam consists of 6 pages including this front page. Ground Rules 1. No calculator is allowed. 2. Show your work for every problem. A correct answer without any justification will receive no credit. 3. You may use one 4-by-6 index card, both sides. Score 1 10 2 10 3 10 4 10 5 10 Total 50 1

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1. (a) Solve y - 4 y + 4 y = 0 . The characteristic equation r 2 - 4 r + 4 = 0 gives r = 2. Thus y = C 1 e 2 t + C 2 te 2 t . (b) Solve y + y + y = 0 The characteristic equation r 2 + r + 1 = 0 gives r = - 1 ± i 3 2 Thus y = C 1 e - 1 / 2 t cos( 3 2 t ) + C 2 e - 1 / 2 t sin( 3 2 t ) . 2
2. Consider t 2 y - 4 ty + 6 y = 0. First verify that y = t 2 is a solution. Then find the general solution by the method of reduction of order. t 2 ( t 2 ) - 4 t ( t 2 ) + 6 t 2 = 2 t 2 - 8 t 2 + 6 t 2 = 0 . Thus y = t 2 is indeed a solution. Now let another solution be y = ut 2 . Then y = u t 2 + 2 ut and y = u t 2 + 4 u t + 2 u . Hence if we plug-in those to the given differential equation, we get t 2 y - 4 ty + 6 y = t 2 ( u t 2 + 4 u t + 2 u ) - 4 t ( u t 2 + 2 ut ) + 6 ut 2 = u t 4 .

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