This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 1 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 ) See Figure 1. 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 2 1 3 1 a . (i) The system has a unique solution if the determinant is not zero. det( M ) = 5 2 a + 1 ( 1) 1 + 3 3 1 5 ( 1) 1 1 3 a 3 2 1 so det( M ) = 7 a + 7 . There will be a unique solution if a 6 = 1. (ii) To have an infinite number of solutions, we first need a = 1, so that the matrix is noninvertible. Then the system becomes 5 x + 3 y + z = 1 x + 2 y + z = 2 3 x y z = b Subtracting the first line from the second and adding the first line to the third, we have 5 x + 3 y + z = 1 4 x y = 8 x + 2 y = b + 1 Adding twice the second row to the third, we have 5 x + 3 y + z = 1 4 x y = 0 = b + 1 Thus, the system will have an infinite number of solutions if a = 1 and b = 1. (iii) From the previous calculations, if a = 1 and b 6 = 1, the system has no solutions. 2 Question 4. [6 points] (a) The eigenvalues of A are the numbers where the polynomial det( A I ) equals 0. So lets evaluate that determinant. A I = 5  2 1 3 , so det( A I ) = (5 )(3 ) 1( 2) = 2 8 + 17 . The roots of this polynomial are = ( 8) p ( 8) 2 4 * 17 2 = 8  4 2 = 4 i, as claimed. (b) The eigenvector corresponding to an eigenvalue is a nonzero solution v to ( A I ) v = . Here is 4 + i , so A I = 1 i 2 1 1 i ....
View
Full
Document
This note was uploaded on 03/19/2011 for the course MAT 1332 taught by Professor Munteanu during the Spring '07 term at University of Ottawa.
 Spring '07
 MUNTEANU
 Derivative

Click to edit the document details