test2 - Math 250A (Kennedy) - Exam 2 — Fall ’07 SHOW...

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

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 250A (Kennedy) - Exam 2 — Fall ’07 SHOW YOUR WORK. Correct answers with no work will get no credit. 1. (16 points total) Consider the differential equation 2;“ 9%; arc Wolfe? @“yb (vs NT a (b) The function = 0 is a solution of this differential equation. Show that your solution in (a) intersects this solution and explain Why this does not contradict the existence—uniqueness theorem. .50 («+11% 707mm (01] c9 07‘ :9 ~Z;F so W ’j‘M/v 80(uILG—J tn/LerJCC Q ~ g 1— %" ' La} defl‘“C€l (-2, o ). u 2 5; a hr a 4 0/ 5-0 Meopfm (4/06, AJ7L +9 W (.h;/‘L\fi’ @an-ftzn (oz, 0). 2. (20 points) Consider the differential equation % 2 y(2 — (a) Find a so that the following is a solution. 2 y(x)=1+e_az 37%)“ z“- e \q¢ 0+ eflwj L 2 1 9 We-” (W New.) , 2(2-+21VW”—2) ¢€*” ((6 6”)" b (“96’”); \464 la": CI; 3'97 (b) Which of the following is a symmetry of the dif. eq. (i) horizontal translation, i.e., if is a solution then y(:z: — c) is too. yer (ii) vertical translation, i.e., if is a solution then + c is too. N0 (0) Find the solution through the point (1, 1) Hint: Note that the solution in part (a) passes through (0,1). (LT 7'(?) > [ :Pt? 1. (43f ('5 a ,Ca(u+~;"‘ 0M1 2/0/3 I_ i “Y VC/ z! qC/O a e- {0(sfiam, CA’DJC (L 5‘0 9”. 9(O/fi’8 {0 {(l): 7(lfiol {a 2 /£7" Cr/fi “(WI~ ytt~d : 3. (18 points) Consider the differential equation, dy a—y The parameter b can be any nonzero number. Find the equilibrium solutions and determine if they are stable, unstable or semistable. (2 — ey/b) a {2— a”) :0 ' ‘54; . (v.10 at C = 2 (etcy 2;— L 3/1, 7/; “L 2 2’6 .» ‘9 6 é > 2—! #0 P I O 1 alt-M‘J “(NHL/e— :— .— Qfl.‘ L <0 L921: 1'1 al‘fl‘?’ ‘faééa 4. (16 points) For the differential equation dy _ 95+?! ._ 1 dx 6 (a) Where are the solution curves increasing and where are they decreasing? I w 7 a ,t‘ e D / (=2) 1‘ --r (0 07" e 7‘ I C. / ( / (b) Where are the solution curves concave up and where are they concave down? 7/ 4M} c,1 “KI‘EL «T9 °fi=6 (/+ 212226 ((+6 x w C v/Cew/ >0 and”, f6 a(u’¢y, Oakc ave tic/a ___#_______________________ 5. (12 points) For this question, you need not show any work. No partial credit will be given on this one. Let y(m) be a solution of the differential equation dy _ 1 d3? ” (y - 002 (a) Let @(m) = Is :1] a solution? +1 /\/o (b) Let @(w) Is y a solution? (e) Let 27(33) = —y(—:1:). Is 3 a solution? 6. (15 points) In the homework you solved the differential equation y’ = %(1 — 3/2) with the initial condition y(()) = O and found, hopefully, em—l y($)_ex+l Now consider the differential equation f’ = 4 — f2 with the initial condition f (0) = 0. Find the solution. Hint: There are two ways to do this. You can ignore the above formula for y(a:) and just solve the dif eq for f. Of you can use scaling. Th7 %(7/; a ;Cé4¢] /({’>=/= «L {(4%) = at $0» flw) , m 11 um 4r — ,ZL )2 abs, :3 ‘1’ 2} “Al/ 59% 4,724 9 gm: 13C”) 2 2. :9,“ ...
View Full Document

This note was uploaded on 03/21/2012 for the course MATH 250a taught by Professor Kennedy during the Fall '11 term at University of Florida.

Page1 / 5

test2 - Math 250A (Kennedy) - Exam 2 — Fall ’07 SHOW...

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