ma241exam2-c-key

ma241exam2-c-key - Spring 2007 Therkelsen MA 241 Exam 2(C...

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Spring 2007 Therkelsen MA 241 - Exam 2 (C) KEY 1. Find the length of the arc of the given curve, over 0 t 1. y = 1 2 t 2 +8 ,x = 1 9 (9+6 t ) 3 2 Solution: dy dt = t and dx dt = 1 6 t ) 1 2 (6)=(9+6 t ) 1 2 . Using the formula for arc length, we end up with the following L = 1 Z 0 q t 2 +((9+6 t ) 1 2 ) 2 dt = 1 Z 0 p t 2 +9+6 tdt = 1 Z 0 p ( t +3) 2 dt = 1 Z 0 t +3 dt = 1 2 t 2 t 1 0 = ² 1 2 ³ (0) = 7 2 2. Find the values of r for which the function y = e rx satis±es the di²erential equation. y 00 4 y 0 5 y =0 1
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Solution: If y = e rx then y 0 = re and y 00 = r 2 e . Given the diFerential equation, we have r 2 e 4 5 e =0 e ( r 2 4 r 5 )=0 e ( r 5)( r + 1 Since e > 0, the solutions of the equation are r = 5 and r = 1. 3. Solve the diFerential equation. (5 + t 2 ) y 0 = ty 2 Solution: (5 + t 2 ) y 0 = ty 2 1 y 2 dy dt = t 5+ t 2 Z 1 y 2 dy = Z t t 2 dt 1 y = 1 2 Z 1 u du 1 y = 1 2 ln | u | + C 1 y = 1 2 ln ¬ ¬ t 2 ¬ ¬ + C y = 2 ln | t 2 | + C y = 2 ln | t 2 | + C 4. ±ind the solution of the diFerential equation that satis²es the given initial condition.
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This note was uploaded on 01/25/2011 for the course MA 241 taught by Professor Mccollum during the Spring '08 term at N.C. State.

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ma241exam2-c-key - Spring 2007 Therkelsen MA 241 Exam 2(C...

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