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The University of British Columbia
MATH 215
Midterm Exam I – February 2010
Name
Student Number
Signature
This exam consists of 4 questions worth 10 marks each. No notes nor calculators.
Problem max score
1.
10
2.
10
3.
4.
10
10 total 40 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 .

February 2010 Math 215 Midterm 1 Page 2 of 5
(10 points) 1.
(a) Using the local existence theorem for first order equations (Theorem 2.4.2), decide if the equation d dt y = t 1 / 2 + ( y − 1) 2 , y (0) = 1 , has exactly one solution near ( t
0
, y
0
) = (0 , 1). Explain your answer.
Answer . (3pt) YES.
The function f ( t, y ) = t 1 / 2 + ( y − 1) 2 and its partial derivative ∂f /∂y = 2( y − 1) are both continuous near ( t, y ) = (0 , 1) for t ≥ 0. Thus there is a unique solution for t ≥ 0 by the unique existence theorem.
Note: You get full marks if you say “no solution because f ( t, y ) is not defined for t < 0.”
Common errors: It makes no sense to say f 0 = 1
2 t − 1 / 2 + 2( y − 1), or call f by dy/dt .
(b) Repeat part (a) for d dt y = t 2 + ( y − 1) 1 / 2 , y (0) = 1 .
Answer . (3pt) Undecidable.
The function f ( t, y ) = t 2 + ( y − 1) 1 / 2 has partial derivative ∂f /∂y = continuous near ( t, y ) = (0 , 1). Thus Theorem 2.4.2 is not applicable.
1
2
( y − 1) − 1 / 2 which is not
Note: Many of you answered “Since ∂f /∂y is not continuous at y = 0, there does not have a solution” or “there are many solutions.” This is not correct. We cannot draw any conclusion from Theorem 2.4.2 since it is not applicable.
(c) Decide the interval of existence for the equation
(9 − t 2 ) y 00 + (sin t ) y 0 + (ln t ) y = e t , y (1) = 1 , y 0 (1) = 2 .
Answer . (4pt) (0 , 3).
The interval has to contain t = 0. The equation can be rewritten as y 00 +
(sin t )
(9 − t 2 ) y 0 +
(ln t )
(9 − t 2 ) y = e t
(9 − t 2 )
.
The solution exists as long 9 − t 2 is not zero and (ln t ) is defined, i.e., t > 0. Thus the interval is the largest one with t > 0 and contains t = 1 but not 3, i.e., (0 , 3).

February 2010 Math 215 Midterm 1 Page 3 of 5
(10 points) 2.
The population of mosquitoes in a certain area increases at a rate proportional to the current population and, in the absence of other factors, the population becomes e times the original population every 10 days. There are 200,000 mosquitoes in the area initially, and predators
(birds, bats, and so forth) eat 20,000 mosquitoes per day. Determine the population of mosquitoes in the area at any day.
Answer . This is revised problem 13 of § 2.3.
Denote the population at day t by P ( t ).
Step 1. In the absence of predators, the population satisfies d dt
P = rP for some growth rate r > 0. Its solution is P ( t ) = P (0) e rt . From P (10) = eP (0), we get e 10 r = e . Thus r =
1
10
.
Step 2. With predators, the population satisfies d dt
P = rP − k, P (0) = 200 , 000 , k = 20 , 000 .
It has integrating factor e − rt , and
( P e − rt ) 0 = − ke − rt ,
P ( t ) e − rt
P ( t ) =
= k r k e − rt r
+ C,
+ Ce rt .
Using P (0) = 10 k and r = 1 / 10, we get C = 0 and
P ( t ) = 10 k = 200 , 000 for all t .
Common error: Solution formula in Step 1 is not the solution for Step 2, and should not be plugged into d dt
P = rP − k .

(10 points)
February 2010
3.
Solve
Math 215 Midterm 1 dy dx
3 x 2 y + y 2
= −
2 x 3 + 3 xy
, y (1) = − 2.
Answer . This is problem 31 on page 133.
Rewrite as
M dx + N dy = 0 , M = 3 x 2 y + y 2 , N = 2 x 3 + 3 xy.
Since
M y
= 3 x 2 + 2 y, N x
= 6 x 2 + 3 y, the equation is not exact. An integrating factor µ should satisfy
( M µ ) y
= ( N µ ) x
.
Try µ = µ ( x ) independent of y . we get M y
µ = N x
µ + N µ x or
µ x
µ
=
M y
− N x
N
=
− 3 x 2 − y
2 x 3 + 3 xy which depends on y , which is not impossible for µ = µ ( x ).
Try µ = µ ( y ) independent of x . we get M y
µ + M µ y
= N x
µ or
µ y
µ
=
− M y
+
M
N x
=
3 x 2 + y
3 x 2 y + y 2
=
1 y
, which is an ODE in variable y . Integrating we get ln µ = ln y + C and we can take
µ = y.
The equation M ydx + N ydy = 0 is exact. Let
Z Z
ψ = M ydx = 3 x 2 y 2 + y 3 dx = x 3 y 2 + xy 3 + h ( y ) .
Page 4 of 5
Thus
ψ y
= 2 x 3 y + 3 xy 2 + h 0 ( y ) = N y = 2 x 3 y + 3 xy 2 , and h 0 ( y ) = 0. We can take h = 0 and
ψ ( x, y ) = x 3 y 2 + xy 3 .
At the point ( x, y ) = (1 , − 2) we have ψ = 4 − 8 = − 4. Thus the solution y = y ( x ) is given implicitly by x 3 y 2 + xy 3 = − 4 .

February 2010 Math 215 Midterm 1
(10 points) 4.
Solve y 00 + 2 y 0 + 5 y = 0 , y (0) = 3 , y 0 (0) = 1.
Answer . Characteristic equation r 2 + 2 r + 5 = 0 .
Two roots r
1
= − 1 + 2 i, r
2
= − 1 − 2 i.
General solution y ( t ) = c
1 e − t cos 2 t + c
2 e − t sin 2 t.
y 0 ( t ) = c
1 e − t ( − cos 2 t − 2 sin 2 t ) + c
2 e − t ( − sin 2 t + 2 cos 2 t ) .
Match boundary conditions c
1
+ 0 = 3
− c
1
+ 2 c
2
= 1
Thus c
1
= 3, c
2
= 2, and y ( t ) = 3 e − t cos 2 t + 2 e − t sin 2 t.
Alternative solution.
Let y ( t ) = c
3 e ( − 1+2 i ) t + c
4 e ( − 1 − 2 i ) t . Then
3 = c
3
+ c
4
, 1 = c
3
( − 1 + 2 i ) + c
4
( − 1 − 2 i ) .
One can solve c
3
= 3
2
− i , c
4
= 3
2
+ i , and thus y ( t ) = Re 2(
3
2
− i ) e ( − 1+2 i ) t = 3 e − t cos 2 t + 2 e − t sin 2 t.
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