Fall 2006 - Takeda's Class - Exam 2

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

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Unformatted text preview: Math 20D (Introduction to Differential Equations) Midterm 2 Solutions November 13, 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 15 3 10 4 10 5 15 Total 50 1 1. (a) Solve y 00- 6 y + 9 y = 0 . The characteristic equation r 2- 6 r + 9 = 0 gives r = 3. Thus y = C 1 e 3 t + Cte 3 t . (b) Solve y 00 + 3 y = 0 The characteristic equation r 2 + 3 = 0 gives r = ± i √ 3. Thus y = C 1 cos √ 3 t + C 2 sin √ 3 t 2 2. Consider ( t- 1) y 00- ty + y = 0 for t > 1. First verify that y = e t is a solution. Then find the general solution by the method of reduction of order. Since ( t- 1)( e t ) 00- t ( e t ) + e t = ( t- 1- t + 1) e t = 0, y = e t is indeed a solution. Now we have to find a function u so that Y = ue t is another solution. Note that Y = ( ue t ) = u e t + ue t Y 00 = ( ue t ) 00 = ( u e t + ue t ) = u 00 e t + 2 u e t + ue t Therefore ( t- 1) Y 00- tY + Y = ( t- 1)( u 00 e t + 2 u e t +...
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Fall 2006 - Takeda's Class - Exam 2 - Math 20D(Introduction...

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