The Predator-Prey System
(Lotka-Volterra equations) If x is the prey species and y the predator, then the Lotka-Volterra equations are
x = x(a y ),
y = y (c + x)
where a, c, , and are positive constants. For the graphs that follow we take a = 1.4, c = 2.0
642:527
SOLUTIONS: ASSIGNMENT 7
FALL 2009
NOTE: A. There is a Maple worksheet on the Assignments and solutions web page which shows the phase plane portraits for problems 2(b) and 2(k). B. Vectors are denoted either by boldface or, for Greek letters, unde
642:527
SOLUTIONS: ASSIGNMENT 6
FALL 2009
NOTE: A. There is a Maple worksheet on the Assignments and solutions web page which shows the phase plane portraits for some of these problems. I do not know how to have Maple put arrows on curves, so for many of
642:527
SOLUTIONS: ASSIGNMENT 5
FALL 2009
Some of these solutions were written by Professor Dan Ocone. 5.5.1 (a), (d): See solutions in text. (b) f (t) = H (t)et H (t 1)et . The rst H (t) is in a sense not needed, since in discussing the Laplace transform
642:527
SOLUTIONS: ASSIGNMENT 4
FALL 2009
Some of these solutions were written by Professor Dan Ocone. 5.2.1 A function f has exponential order as t if there exist constants K 0, T 0, and a constant c, such that |f (t)| Kect for all t T . (4.1)
A straight
642:527
SOLUTIONS: ASSIGNMENT 3
FALL 2007
4.3.6 (e) This is an Euler equation and one can look for its solution in the form xr without need for the full series. Plugging this into the equation x2 y + xy y = 0 leads to xr (r (r 1) + r 1) = 0. Hence xr will
642:527
SOLUTIONS: ASSIGNMENT 2
FALL 2009
Some of these solutions were written by Professor Dan Ocone. 4.3.1 (a) The answer in the text seems sucient. (b) Rewrite the equation as y [(cos x)/x]y +[5/x]y = 0. Then p(x) and q (x) are analytic at all points e
4 cq Sam p4 Sxm p4 p am q qq "pS4am p vm % "pS4tam 4 p am % 4 4 p am
vkUUj|q|v!kyQj hi e l e r x r h fi 4
p 4 m l ei h f e ri f r 6 p m qqd!3qrVkjkikUkUFvqQ7 Q|qe r x e hi hi hi Uykekks3Q UUxcfw_ h i f l e h r p m p iki~izijzex7farS4 p p 4 m p m m p
642:527
REVIEW EXAM 2
FALL 2009
1. (a) Find the general solution of z = Az, where z =
x 2 3 and A = . Be sure you y 2 5 actually give this solution (which should involve two free constants). (b) On the axes below give a careful drawing of the phase plane
642:527
SOLUTIONS: ASSIGNMENT 8
FALL 2009
Section 9.10: 2 (a) See text solution. (c) With u = (3, 0, 1, 4, 1) we have 1 , u = 5/ 5, 2 , u = 6/ 6, 3 , u = 4, so the best e e e approximation is
3
Section 9.9: 4 (a) See text solution. (e) The formula is (23)
642:527 EXPANSIONS IN ORTHOGONAL BASES Vector spaces
FALL 2009
We will use without much further comment the idea of a vector space. Basically, a vector space is a set of vectors (these vectors will in fact often be functions) with the property that a line
642:527
FALL 2009
SUMMARY OF THE METHOD OF FROBENIUS Consider the linear, homogeneous, second order equation: y + p(x)y + q (x)y = 0. Suppose that x = 0 a regular singular point:
(1)
xp(x) =
n=0
pn x ,
n
| x| < R 1 ,
x q ( x) =
n=0
2
qn xn ,
| x| < R 2
642:527:01
APPROXIMATE SYLLABUS
FALL 2009
This syllabus is tentative; elements such as scheduling of exams and coverage on each exam may change. All changes will be posted on the class web site. Section numbers refer to Advanced Engineering Mathematics (2
642:527
SOLUTIONS: ASSIGNMENT 13
FALL 2009
Section 19.2: 5. (a) The wire loop is treated as massless, so any force on it would cause innite acceleration: there can be no (net) force on the loop. A nonzero ux (L, t) would lead to a nonzero vertical force,
642:527
SOLUTIONS: ASSIGNMENT 12
FALL 2009
Section 17.10: 2. If Re a > 0 then limx e(ai)x = 0 for real, so F cfw_H (x)eax =
0
H (x)eax eix dx =
e(ai)x dx =
e(ai)x a i
0
=
1 . a i
3. If a > 0 then all the integrals below converge: F cfw_e
a|x|
=
0
e
a
642:527
SOLUTIONS: ASSIGNMENT 11
FALL 2008
Section 17.7: 7. Let us rst consider > 0 and then write the solution of the ODE as y = A cos x + B sin x. Then y = A sin x + B cos x the boundary conditions y (0) y (1) = 0 and y (0) + y (1) = 0 become A (A cos +
642:527
SOLUTIONS: ASSIGNMENT 10
FALL 2009
Section 17.7:1. In all sections of this problem the equation is y + y = 0; comparing with the standard Sturm-Liouville form (1.a) we see that p(x) = 1, q (x) = 0, and w(x) = 1. The solutions of the dierential equ
642:527
SOLUTIONS: ASSIGNMENT 9
FALL 2009
Section 17.3: 4 (g) f (x) = | sin x| has period (since sin(x + ) = sin x) so its Fourier series will have the form
FS f = a0 +
n=1
[an cos 2nx + bn sin 2nx].
Since f (x) is even (| sin(x)| = | sin x| = | sin x|) t
642:527 (16)
REVIEW EXAM 1
FALL 2009
1. (a) Find the Laplace transform Y (s) of the solution y (t) of the initial value problem y + 2y + 5y = C (t ), where C is a constant. (b) Find y (t) by taking the inverse Laplace transform of Y (s). (c) Find a value
642:527
PREREQUISITE QUIZ: SOLUTIONS 2n x3n . n2 n=0
Fall 2009
1. Find the radius of convergence of the power series
Solution: We use the ratio test: if bn = 2n x3n /n2 is a typical term of the series then
n
lim
2n+1 x3(n+1) /(n + 1)2 2|x|3 n2 bn+1 = lim
642:527
ASSIGNMENT 10
FALL 2009
No problems from this assignment will be collected. These exercises are just for you to look at simple Sturm-Liouville problems. Some solutions will be posted. Section 17.7: 1 (a), (b), (c), (d), (e), (f); 2 Comments, hints
642:527
ASSIGNMENT 7
FALL 2009
Turn in starred problems Tuesday 10/27/2009. Section 7.4: 2 (a), (b)*, (f)* Section 7.4: 7 Section 7.5: 4* Problem 7.A* Two interacting populations x(t), y (t) are described by the equations x = (1 x)x , y = (3 y x)y . See i
642:527
ASSIGNMENT 6
FALL 2009
CHANGE IN DUE DATE OF THIS ASSIGNMENT: Turn in starred problems Thursday 10/15/2009. Be sure to read the instructions below. Section 7.2: 1; 4 (a), (b)*; 5 (a) (c) (f); 10 Section 7.3: 1 (a), (c)*, (h); 9 (a), (g), (h)*, (j)