{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mid2-2010

# mid2-2010 - Question 1[4 points Consider the dierential...

This preview shows pages 1–3. Sign up to view the full content.

Question 1. [4 points] Consider the differential equation dx dt = x - a 2 x 3 . (a) Find all equilibria. (b) Determine the stability of each equilibrium point. (c) Draw the phase-line diagram for the differential equation. Solution. a) Equilibria occur when f ( x ) = x - a 2 x 3 = 0. Factoring, we have x (1 - a 2 x 2 ) = 0 so x = 0 , ± 1 a . Version 1 : a = 2, so the answer is x = 0 , ± 1 2 . Version 2 : a = 3, so the answer is x = 0 , ± 1 3 . Version 3 : a = 6, so the answer is x = 0 , ± 1 6 . b) Differentiating, we have f 0 = 1 - 3 a 2 x 2 f 0 (0) = 1 > 0 f 0 ( ± 1 ) = 1 - 3 a 2 1 < 0 0 1 a - 1 a Figure 1: 1(c): x = 0 is unstable; the other two points are stable. Question 2. [6 points] Consider the equations Version 1 : 2 x 3 - 8 x 2 + 11 x - 6 = 0 . Version 2 : 2 x 3 - 10 x 2 + 15 x - 9 = 0 . Version 3 : 2 x 3 - 12 x 2 + 19 x - 12 = 0 . 1

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

View Full Document
(a) Show that x 1 = 2 ( x 1 = 3) [ x 1 = 4] is a solution of the equation. (b) Use long division to find x 2 and x 3 , the other two solutions. (c) Calculate ¯ x 2 /x 3 , where ¯ w is the complex conjugate of w . (d) Find the four roots of x 4 = c 2 in the form a + ib , where a and b are real. (e) Express each root in part (d) in the form re with r > 0. Solution. a) Version 1 : When x = 2, we have 2(2 3 ) - 8(2 2 ) + 11(2) - 6 = 0. Version 2 : When x = 3, we have 2(3 3 ) - 10(3 2 ) + 15(3) - 9 = 0. Version 3 : When x = 4, we have 2(4 3 ) - 12(4 2 ) + 19(4) - 12 = 0. b) Long division factors the equation as ( x - a )(2 x 2 - 4 x + 3) = 0 . Thus, using the quadratic formula, the other two solutions satisfy x = 1 ± p 16 - 4(2)(3) 4 = 4 ± 8 i 4 = 1 ± i 2 (You might notice some similarity between this and an assignment question!) c) It actually doesn’t matter which solution is x 2 and which is x 3 , as the answer will be the same. If x 2 = 1 - i 2 , then ¯ x 2 = 1 + i 2 and thus ¯ x 2 x 3 = 1 + i 2 1 + i 2 = 1 d) If x 4 = c 4 , then we have x 2 = c 2 or x 2 = - c 2 . From x 2 = c 2 , we have x = ± c . From x 2 = - c 2 , we have x = ± ci . Version 1 : c = 5, so x = ± 5, ± 5 i .
This is the end of the preview. Sign up to access the rest of the document.
• Spring '07
• MUNTEANU
• SEPTA Regional Rail, 11:11, Eigenvalue, eigenvector and eigenspace, Tier One, Scaled Composites, Scaled Composites White Knight

{[ snackBarMessage ]}

### Page1 / 10

mid2-2010 - Question 1[4 points Consider the dierential...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online