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# midterm2 - Math 108 Second Midterm Examination NAME(Please...

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Unformatted text preview: Math 108 Second Midterm Examination October 24, 2007 NAME (Please print) Page Possible Score 2 10 3 10 4 10 5 10 6 15 7 15 8 10 9 10 10 10 Total 100 Instructions: 1. Do all computations on the examination paper. You may use the backs of the pages if necessary. 2. Put answers inside the boxes (when applicable). 3. Provide arguments for functions, and bounds on integrals. 4. Write equations and inequalities, not disconnected mathematical expressions. 5. Please signify your adherence to the honor code: I, , have neither given nor received unauthorized help on this exam and I have conducted myself within the guidelines of the Duke Community Standard. 1 (10 Points) Score 1. (5 points) What is the general solution for y ( x ) if 4 y 00 ( x ) + y ( x ) = 0 ? Try solutions of the form y ( x ) = e rx . The characteristic polynomial 4 r 2 + 1 = 0 has roots r = ± ı/ 2 so the general solution is y ( x ) = c 1 cos( x/ 2) + c 2 sin( x/ 2). y ( x ) = c 1 cos( x/ 2) + c 2 sin( x/ 2) 2. (5 points) What is the general solution for y ( x ) if y 00 ( x )- 6 y ( x ) + 9 y ( x ) = 0 ? Try solutions of the form y ( x ) = e rx . The characteristic polynomial r 2- 6 r +9 = 0 has double root r = 3 so the general solution is y ( x ) = c 1 e 3 x + c 2 xe 3 x . y ( x ) = c 1 e 3 x + c 2 xe 3 x 2 (10 Points) Score 1. (5 points) What is the general solution for y ( x ) if ( x + 1) 2 y 00 ( x ) + 3( x + 1) y ( x ) + 3 4 y ( x ) = 0 ? Try solutions of the form y ( x ) = ( x + 1) r , corresponding to the substitution x + 1 = e z or z = log( x + 1). The indicial equation 0 = r ( r- 1) + 3 r + 3 / 4 = r 2 + 2 r + 3 / 4 has roots r =- 1 ± 1 / 2 =- 1 / 2 ,- 3 / 2, so the general solution is y ( x ) = c 1 ( x + 1)- 1 / 2 + c 2 ( x + 1)- 3 / 2 . y ( x ) = c 1 ( x + 1)- 1 / 2 + c 2 ( x + 1)- 3 / 2 2. (5 points) What is the general solution for y ( x ) if x 2 y 00 ( x )- 3 xy ( x ) + 4 y ( x ) = 0 ? Try solutions of the form y ( x ) = x r . The indicial equation 0 = r ( r- 1)- 3 r + 4 = r 2- 4 r + 4 has double root r = 2, so the general solution is y ( x ) = c 1 x 2 + c 2 x 2 ln x . y ( x ) = c 1 x 2 + c 2 x 2 ln x 3 (10 Points) Score 1. (5 Points) What is the general solution for y ( x ) if x 2 y 00 ( x ) + 3 xy ( x ) + 5 y ( x ) = 0 ?...
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midterm2 - Math 108 Second Midterm Examination NAME(Please...

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