Be sure this exam has 12 pages including the cover
The University of British Columbia
Final Exam – December 2007
Mathematics 215/255, Ordinary Differential Equations
Name
Student Number
Circle Section: 101 Phan
Signature
102 Burghelea
103 Tsai
104 Flowers
105 Brydges
This exam consists of 8 questions worth 100 marks in total. No aids are permitted.
Problem
1.
max score
10
2.
10
3.
10
4.
10
5.
15
6.
15
7.
15
8.
15
total
100
score
1. Each candidate should be prepared to produce his library/AMS card upon request.
2. Read and observe the following rules:
No candidate shall be permitted to enter the examination room after the expiration of one half hour, or to leave
during the first half hour of the examination.
Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities
in examination questions.
CAUTION - Candidates guilty of any of the following or similar practices shall be immediately dismissed from the
examination and shall be liable to disciplinary action.
(a) Making use of any books, papers or memoranda, other than those authorized by the examiners.
(b) Speaking or communicating with other candidates.
(c) Purposely exposing written papers to the view of other candidates. The plea of accident or forgetfulness
shall not be received.
3. Smoking is not permitted during examinations.
December 2007
Math 215/255 Final Exam, All Sections
Page 2 of 12
(10 points) 1. A bath contains water. Let the depth of water at time t be y = y(t) ins. At time t = 0 the
plug is pulled and water exits via the drain at the rate of 2y inches per minute.
(a) The bath is filled to an initial depth of 10 inches. Find the depth y as a function of time.
dy
dt = −2y
y = y0 e−2t = 10e−2t
(b) Same problem as in the previous part except that at time t = 0 the tap is turned on and
water enters the bath at rate 3 inches per minute.
dy
dt = −2y + 3
dy 2t
2t
2t
dt e +2ye = 3e
d
e2t y = 3e2t
hdt
is=t
h it
e2s y(s)
= 32 e2s
0 i
s=0
h
3 2t
2t
e y(t) − 10 = 2 e − 1
y(t) = 10e−2t +
3
2
h
1 − e−2t
i
December 2007
Math 215/255 Final Exam, All Sections
Page 3 of 12
(10 points) 2. Consider
t + ye2ty + ate2ty
dy
= 0.
dt
(a) Determine a such that (1) is exact.
Denote M = t + ye2ty and N = ate2ty .
My = e2ty (1 + 2ty),
Nt = e2ty (a + 2aty)
To be exact we need My = Nt . Thus a = 1.
(b) Solve (1) when a has the value found in the previous part.
φ(t, y) =
Z
M dt =
1 2 1 2ty
t + e + f (y).
2
2
φy = 0 + te2ty + f 0 (y)
We need φy = N . Thus f 0 = 0 and
φ=
or y =
1
2t
ln(c0 − t2 ).
1 2 1 2ty
t + e + c.
2
2
(1)
December 2007
Math 215/255 Final Exam, All Sections
Page 4 of 12
(10 points) 3. For the equation y 0 = 1 − x − y with initial condition y(0) = 2 write down the Euler recursion
with h = 0.1 for finding the solution approximately and use it to approximate y(0.2).
Let f (x, y) = 1 − x − y and h = 0.1. We have the recursion formula
y0 = 2,
yk+1 = yk + h · f (kh, yk ),
k = 0, 1, 2, . . .
Thus
y1 = 2 + 0.1 · f (0, 2) = 2 + 0.1 · (−1) = 1.9.
y2 = 1.9 + 0.1 · f (0.1, 1.9) = 1.9 + 0.1 · (−1) = 1.8.
December 2007
Math 215/255 Final Exam, All Sections
(10 points) 4. (a) Find the general solution to y 00 − 2y 0 + 2y = 0.
r 2 − 2r√+ 2 = 0.
r = 2± 24−8 = 1 ± i.
y = Aet cos t + Bet sin t.
(b) Find a particular solution to y 00 − 2y 0 + 2y = et cos t
y = <ve(1+i)t . v solves
(D + 1 + i)2 v − 2(D + 1 + i)v + 2v = 1
D 2 v + 2(1 + i)Dv + 2iv − 2Dv − 2(1 + i)v + 2v = 1
D 2 v + 2iDv = 1
Let w = Dv: Dw + 2iw = 1.
1
w = 2i
1
v = 2i
t.
1 (1+i)t
y = < 2i
te
= 21 tet sin t
(c) Find the general solution to y 00 − 2y 0 + 2y = et cos t.
y = Aet cos t + Bet sin t + 21 tet sin t
Page 5 of 12
December 2007
(15 points) 5. (a) Solve
Math 215/255 Final Exam, All Sections
y 00
+
y0
=
(
t
0
if t ≥ 1
when y(0) = y 0 (0) = 0.
otherwise
(s2 + s)Y (s)
= L {tH1 (t)} = e−s L {t + 1}
1
s+1
1
−s
=e
= e−s 2
+
2
s
s
s
1
Y (s) = e−s 3
s
(t − 1)2
y(t) = H1 (t)
2
⇒
⇒
(b) Find F (s) = L
n
1
√
t
o
√
√
√
, given that L{ t} = 12 πs−3/2 and F (1) = π.
n√ o
t
L
t = L √ = −F 0 (s)
t
⇒
√
1 √ −3/2
πs
= −F 0 (s) , F (1) = π
2
√
⇒ F (s) = πs−1/2 .
(c) Find the inverse Laplace transform of
L
−1
Page 6 of 12
e−2s
(s + 1)2
= H2 (t)L
−1
e−2s
.
(s + 1)2
1
(s + 1)2
(t − 2) = H2 (t)(t − 2)e−(t−2)
December 2007
Math 215/255 Final Exam, All Sections
Page 7 of 12
(15 points) 6. Consider
d
dt
x
y
=A
(a) Find the solution of (2) such that
− 21
,
A=
−1
x(0)
1
=
.
y(0)
0
x
y
x(0)
y(0)
Eigenvalues: λ2 + λ + 45 = 0. λ1 = − 21 + i, λ2 = − 21 − i.
Eigenvector for λ1 :
0
u
−i 1
=
0
v
−1 −i
1
u
.
=
i
v
1
− 2t it
Complex solution: e e
i
0
1
− 2t
+
i
cos t + i sin t
e
1
0
General real solution
cos(t)
sin(t)
− 2t
− 2t
C1 e
+ C2 e
− sin(t)
cos(t)
−0.5t
e
cos(t)
1
x(0)
.
is
=
Solution with
−e−0.5t sin(t)
0
y(0)
(b) Find the solution of (2) such that
−0.5t
e
sin(t)
.
e−0.5t cos(t)
=
0
.
1
1
−
1
2
.
(2)
December 2007
Math 215/255 Final Exam, All Sections
Page 8 of 12
(c) Find the matrix etA .
The fundamental matrix:
X(t) = e
e
At
= X(t)X
−1
− 2t
(0) =
cos(t) sin(t)
− sin(t) cos(t)
e−t/2 cos(t)
e−t/2 sin(t)
−e−t/2 sin(t) e−t/2 cos(t)
.
(d) Draw the phase portrait for (2) including directions of flow.
Noting that Im(λ1 ) 6= 0 and Re(λ1 ) < 0 one can conclude that the trajectories are stable
spirals. In order to figureout whether it is clockwise or counter-clockwise, plug the
→
−
0
initial condition X (0) =
in the initial system and obtain:
1
→
−0
X (0) =
−1/2 1
−1
−1/2
0
1
=
1
−1/2
Therefore, x0 (0) = 1 > 0 and y 0 (0) = −1/2 < 0. Therefore we deal with a clock wise
spiral. The phase portrait is illustrated in the figure below.
2
1.5
1
y
0.5
0
−0.5
−1
−1.5
−2
−2
−1.5
−1
−0.5
0
x
0.5
1
1.5
2
December 2007
(15 points) 7. Consider
d
dt
Math 215/255 Final Exam, All Sections
x
y
=
(a) Find the orbits.
dy
2x+1
dx = 2y+y 3
2y + y 3
2x + 1
Page 9 of 12
dy
= 2x + 1
(2y + y 3 ) dx
1 4
2
y + 4 y + C = x2 + x + 41 = (x + 21 )2 .
q
x = − 21 ± y 2 + 14 y 4 + C
(b) Find and classify equilibrium point(s).
Solve 2y + y 3 = 0 and 2x + 1 = 0
(x0 , y 0 ) = (− 12 , 0)
0 2
Jacobian =
2 0
Eigenvalues: λ = ±2 Saddle point
(c) Draw the local phase plane portrait near the equilibrium point(s).
1
−1
Eigenvectors λ = 2:
; λ = −2:
1
1
2
1
y
0
−1
−2
−3
−4
−2
−1
0
1
x
2
3
4
December 2007
Math 215/255 Final Exam, All Sections
Page 10 of 12
(15 points) 8. For the following questions, fill in the answers in the boxes. No work need to be shown and no
partial credit will be given. Anything written outside of the boxes is ignored.
(a) Convert the equation
dy
d2 y
+ p + qy = 0,
2
dt
dt
y(0) = 1,
y 0 (0) = 2.
into an equivalent first order system of the form
d x1
?
x1 (0)
?
.
=
,
=
?
x2 (0)
?
dt x2
Answer =
d
dt
x1
x2
=
x2
−qx1 − px2
,
x1 (0)
x2 (0)
=
(b) In the method of Picard iteration the equation
solved by a recursion of the form
Z
yn+1 (t) =? +
1
2
dy
dt
.
= y 2 , with initial condition y(0) = 7 is
?
? ds,
y0 (t) =?
?
Find y1 (t).
Answer =
Missing parts
yn+1 (t) = 7 +
Z
t
yn (s)2 ds,
y0 (t) = 7
0
We have
y1 (t) = 7 +
(c) Is the solution to
dy
dt
Z
t
49ds = 7 + 49t
0
= y 2/3 with y(0) = 1 unique?
Answer =
YES
(d) Is the solution to
dy
dt
= y 2/3 with y(0) = 0 unique?
Answer =
NO
December 2007
Math 215/255 Final Exam, All Sections
Page 11 of 12
(e) The Wronskian of two solutions of y 00 + 3y 0 + 2y = 0 is proportional to eat . Find a.
Answer =
Z
W [y1 , y2 ] = c exp − p(t)dt
Here p(t) = 3, so a = −3.
December 2007
Math 215/255 Final Exam, All Sections
Table of Laplace transforms
f (t)
C
eat
tn , n positive integer
tp , p > −1
sin(at)
cos(at)
sinh(at)
cosh(at)
δ(t − c)
F (s) = L[f (t)]
C
s
1
s−a , s > a
n!
,s > 0
sn+1
Γ(p+1)
,
sp+1
a
,
s2 +a2
s
s2 +a2 ,
a
,
s2 −a2
s
,
s2 −a2
e−cs
s>0
s>0
s>0
s > |a|
s > |a|
eat · sin(bt)
b
,
(s−a)2 +b2
s>a
eat · cos(bt)
s−a
,
(s−a)2 +b2
s>a
tn · eat , n positive integer
n!
,s
(s−a)n+1
>a
Hc (t)
Hc (t)f (t − c)
Rt
0
ect · f (t)
f (t − τ )g(τ )dτ
f (n) (t)
(−t)n f (t)
e−cs
s ,
s>0
e−cs F (s)
F (s − c)
F (s)G(s)
sn F (s) − sn−1 f (0) − ... − f (n−1) (0)
F (n) (s)
Page 12 of 12