midterm2

midterm2 - Math 108 Second Midterm Examination October 24,...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 = / 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 ?...
View Full Document

Page1 / 11

midterm2 - Math 108 Second Midterm Examination October 24,...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online