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Unformatted text preview: Question 1. [4 points] a ) An equilibrium point x * of the differential equation dx dt = F ( x ) satisfies F ( x * ) = 0. We have F ( x * ) = 25 x * ( x * ) 3 = x * (25 ( x * ) 2 ) , so the equilibria are , 5 , 5 . b ) We have F ( x ) = 25 3 x 2 . The derivative test gives F (0) = 25 > = 0 the equilibrium is unstable F (5) = 50 > 0 = 5 the equilibrium point is stable and F ( 5) = F (5) = 50 > 0 =  5 the equilibrium point is stable. We could also draw the graph of the function and determine the stability using the regions of increase or decrease. c ) / >  5 < o / > 5 < o Question 2. [6 points] a ) It is easy to verify that f (2) = 8 24 + 26 10 = 0. b ) Dividing f ( x ) by x x 1 = x 2 gives g ( x ) = f ( x ) x 2 = x 2 4 x + 5. g ( x ) = 0 implies x 2 / 3 = 1 2 (4  4) = 2 i . c ) x 2 x 3 = (2 i )(2 + i ) = 4 2 i + 2 i i 2 = 5 . d ) x 2 x 3 = 2 i 2 + i = (2 i )(2 i ) (2 i )(2 + i ) = 4 2 i 2 i + i 2 5 = 3 5 4 5 i. e ) (3 + i )(3 + i ) = 9 + 3 i + 3 i i 2 = 8 + 6 i. f ) Using the relationship e i = cos( ) + i sin( ) we find e i = cos(1) + i sin(1) . Question 3. [5 points] (a) The determinant of B is 1 det( B ) = 1 4 0 + 2 1 c + 3 5 1 1 5 2  3 4 c so det( B ) = 10 c 5 . The matrix B is invertible is and only if det( B ) 6 = 0, ie c 6 = 1 / 2. (b) The associated augmented matrix is M = 5 3 1 1 1 2 1 2 3 1 a b (i) The system has a unique solution if the determinant is not zero....
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This note was uploaded on 09/17/2011 for the course MAT 1332 taught by Professor Munteanu during the Winter '07 term at University of Ottawa.
 Winter '07
 MUNTEANU
 Derivative

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