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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/ 50 Meopfm (4/06, AJ7L +9 W (.h;/‘L\ﬁ’ @anftzn (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.
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“(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 altM‘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
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 Fall '11
 Kennedy
 Calculus

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