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Solution to Midterm Test 2 (version A)
MAT 1322D, Fall 2016 Total = 20 marks
Part I. Multiple-choice questions (3 4 = 12 marks)
Version A1: CDAE Version A2: DABD
- Suppose Euler's method with step size h = 0.05 is used to estimate y (0.1), where y ( t ) is the solution to the initial-value problem y' = (2 t 1) y , y (0) = 1. Which one of the following values is closest to the result that you obtained?
Version A1: (A) 0.95; (B) 0.93; (C) 0.91; (D) 0.89; (E) 0.87. Version A2: (A) 0.85; (B) 0.87; (C) 0.89; (D) 0.91; (E) 0.93.
Solution. The iteration formula is yn +1 = yn + h (2 tn – 1) yn with t (^) 0 = 0, and y 0 = 1.
n tn yn 0 0 1 1 0.05 1 + 0.05 (2 0 − 1) 1 = 0. 2 0.1 0.95 + 0.05 (2 0.05 – 1) 0.95 = 0.
y (0.1) 0.91.
- Suppose salted water of concentration 5 g / m^3 is added to a reservoir of volume 1000 m^3 at a rate 2 m^3 / minute. Assume the water in the reservoir is well mixed and the same amount of mixed water is removed from the reservoir. Let Q ( t ) be the quantity, in grams, of salt in the reservoir at time t. The differential equation that Q ( t ) satisfies is
Version A1:
(A) Q' = 10 Q − 0.002 Q^2 ; (B) Q' = 5 – 0.002 Q ; (C) Q' = 10 – 0.005 Q ; (D) Q' = 10 – 0.002 Q; (E) Q' = 5 Q – 0.005 Q^2.
Version A2:
(A) Q' = 10 – 0.002 Q; (B) Q' = 5 Q – 0.005 Q^2 ; (C) Q' = 10 Q − 0.002 Q^2 ; (D) Q' = 5 – 0.002 Q ; (E) Q' = 10 – 0.005 Q.
Solution. r in = 2 5 = 10. r out = 2 1000
^ Q = 0.002 Q. The equation is Q' = 10 – 0.002 Q.
- The sum of the series
2 1 0
n n n n n
^
is
Version A1:
(A)^25
; (B)^55
; (C)^33
; (D)^35
; (E)^28
Version A2:
(A)^35
; (B)^25
; (C)^55
; (D)^28
; (E)^33
Solution.
2 1 2 1 0 0 0
n n n n n n n n^ n n^ n n
. Both series on the right-hand side are
geometric series. The first series has first term 1, and common ratio^4 5
, and the second series
has first term 3 and common ratio −^3 5
. The sum of the series is S = 1 3 5 15 25 1 4 / 5 1 3/ 5 8 8
- Consider series
2 1 2 2
2 sin n^2 cos
n n n n
. Which one of the following statements is true?
Note that, when n 1, we have 2 n < 2 n + sin^2 n < 3 n and 2 n^2 < 2 n^2 + cos^2 n < 3 n^2. Hence
2 2 2 2 2
2 2 sin 3 3 2 cos 2
n n n n n n n n
when n 1.
Version A1:
(A) Since
2 2 2 2 3/ 2
2 sin 2 2 2 cos 3 3
n n n n n n n
and 3/ 2 3/ 2 1 1
n^3 n^^3 n n
diverges,
2 1 2 2
2 sin n^2 cos
n n n n
diverges.
(B) Since
2 2 2 2 3/ 2
2 sin 2 2 2 cos 3 3
n n n n n n n
and 3/ 2 3/ 2 1 1
n^3 n^^^3 n n
converges,
2 1 2 2
2 sin n^2 cos
n n n n
converges.
(C) Since
2 2 2 2 3/ 2
2 sin 2 2 2 cos 3 3
n n n n n n n
and 3/ 2 3/ 2 1 1
n^3 n^^^3 n n
converges,
2 1 2 2
2 sin n^2 cos
n n n n
diverges.
Part II. Detailed Answer questions (8 marks)
- (4 marks) (A1) Solve the initial-value problem y' = (1 + y )(5 − y ), y (0) = −3.
Solution.^1 (1 )(5 )
dy dt y y
Since^1 1 1 1 , 1 ln^1 (1 )(5 ) 6 1 5 6 5
y (^) t C y y y y y
^ ^
6 1
y (^) K e t y
, where K 1 = e^6 C^ > 0.^16 5
y (^) Ke t y
, K = K 1 0.
By the initial values condition, K = −^1 4
.^1
y (^) e t y
. 4 + 4 y = − 5 e^6 t^ + ye^6 t.
y =
6 6
t t
e e
(A2) Solve the initial-value problem y' = (1 − y )(5 + y ), y (0) = 3.
Solution.^1 (1 )(5 )
dy dt y y
Since^1 1 1 1 , 1 ln^5 (1 )(5 ) 6 1 5 6 1
y (^) t C y y y y y
^ ^
6 1
y (^) K e t y
, where K 1 = e^6 C^ > 0.^56 1
y (^) Ke t y
, K = K 1 0.
By the initial values condition, K = −4.^5 1
y (^) e t y
. 5 + y = −4 e^6 t^ + 4 ye^6 t.
y =
6 6
t t
e e
- (2 2 = 4 marks) Determine whether each of the following series is convergent or divergent. Justify your answer.
Version A1:
a. 2
n n^ ln n
; b.
2
n n
n n
Solution. a. Since function y =^1 x ln x
is positive, decreasing and continuous when x 2, we can
use the integral test.
ln 2 2 ln 2
(^1) lim 1 lim 1 lim ln ln ln ln 2 ln ln
b b x x dx^ b^ x x dx^ b^ udu^ b b
. This
series diverges.
b. Sincelim^1 n 3 1 3
n n
0, this series diverges.
Version A2:
a. 2 2
n n^ (ln^ n )
; b. 2 2
n n
n n
Solution. a. Since function y =^12 x (ln x )
is positive, decreasing and continuous when x 2, we
can use the integral test.
ln 2 2 2 2 ln 2^2
(^1) lim 1 lim 1 lim 1 1 1 (ln ) (ln ) ln ln 2 ln 2
b b x x dx^ b^ x x dx^ b^ u du b b
^ ^
^
This series converges.
b. This is an alternating series. Since 2 3 1
n n
is decreasing and lim 2 0 n 3 1
n n
, this series
converges.
