MA 102
QUIZ 1
MATHEMATICS II
11:00 AM–12:00 NOON
9rd FEB. 2013
IIT GUWAHATI
Instructions
1. To ‘print’ means to write legibly in capital letters.
2. Print your tutorial group:→
T
3. Put your signature in the space provided:→
4. Print your roll number:→
5. Print your name as registered:
6. Do not write anything else on this page.
7. Write your answers in this booklet and only in the space marked for answers. Additional space is
to be found towards the end of the booklet (page no. 6 and 7). Use the last page only for rough
work. The last page (page no. 8) will not be checked.
8. Your writing should be legible and neat. When in doubt, print.
9. Check that you have 8 printed pages. This exam has 4 questions, for a total of 10 points.
10. NO MARKS will be awarded for writing ONLY THE FINAL ANSWER in the space given below
each question.
Space For Recheck Cribs
For Office Use Only
Question:
1
2
3
4
Total
Points:
3
2
2
3
10
Invigilator:
Score:
Grader
Scrutinizer
Rechecker
Version: 1.246
Page 1 of 8
2013 Spring
QUIZ 1
MA 102 : ODE
[3pnts. ] 1. Write the form of yp (x) that one would use in the method of undetermined coefficients for
finding a solution of the non-homogeneous ODE
d3 y
d2 y
dy
= xe−2x cos 3x.
+
4
+ 13
3
2
dx
dx
dx
(You need not compute the coefficients.)
Soln.:
Writing D =
d
,
dx
the given ODE can be written as
(D2 + 4D + 13)Dy = xe−2x cos 3x.
Note that e−2x cos 3x is a solution of the ODE
(D2 + aD + b)y = 0, where − a = (−2 + 3i) + (−2 − 3i) = −4,
b = (−2 + 3i)(−2 − 3i) = 13.
Therefore, xe−2x cos 3x is a solution of the ODE (D2 + 4D + 13)2 y = 0.
Applying the operator (D2 + 4D + 13)2 on both sides of the given ODE, we find that
(D2 + 4D + 13)3 Dy = (D2 + 4D + 13)2 (xe−2x cos 3x) = 0.
Therefore, we can take
yp = [a1 + b1 e−2x cos 3x + b2 e−2x sin 3x]
+c1 xe−2x cos 3x + c2 xe−2x sin 3x + d1 x2 e−2x cos 3x + d2 x2 e−2x sin 3x,
...(2 Mark)
where the bracketed portion can be omitted from yp (x) as that appears in the complementary
function.
...(1 Mark)
Remark: if someone gets only the annhilator correctly, he/she will get 1 mark.
Version: 1.246
Page 2 of 8
2013 Spring
QUIZ 1
MA 102 : ODE
[2pnts. ] 2. Suppose y(x) satisfies the following initial value problem:
dy
+ y tan x = f (x),
dx
y
π 6
(
cosec x 0 < x < π4
.
where f (x) =
π
≤ x < π2
sec x
4
ln 2
= √
3
Determine y( π3 ).
Soln.:
R
Multiplying the given first order linear ODE by the integrating factor e
obtain
[(sec x)y]0 = f (x) sec x.
tan xdx
= sec x, we
...(0.5 Mark)
We now integrate the above equation from
π
3
Z
π
6
to π3 :
0
π
3
Z
[(sec x)y] dx =
f (x) sec xdx
π
6
π π
π π
=⇒ sec
y
− sec
y
=
3
3
6
6
...(0.5 Mark)
π
6
π
4
Z
π
3
Z
cosec x sec x dx +
π
6
sec2 x dx
π
4
...(0.5 Mark)
2 ln2
π π
π π
π
− √ √ = ln tan
− ln tan
+ tan
− tan
3
4
6
3
4
3 3
√
√
= ln 3 + ( 3 − 1)
√
√
π 2 ln 2
=⇒ 2y
−
= ln 3 + ( 3 − 1)
3
3
√
√
π ln 2 ln 3
3−1
=
+
+
.
=⇒ y
3
3
2
2
=⇒ 2y
...(0.5 Mark)
Version: 1.246
Page 3 of 8
2013 Spring
QUIZ 1
MA 102 : ODE
[2pnts. ] 3. Let y1 (x) = x(x + 1), y2 (x) = x(x + 1)2 and y3 (x) = x(x2 + 1) be three solutions of the ODE
d2 y
dy
+ p(x) + q(x)y = g(x)
2
dx
dx
x ∈ (0, 2),
where p(x), q(x) and g(x) are continuous functions on (0, 2). Determine the solution of the
above ODE satisfying the initial conditions y(1) = 2 and y 0 (1) = 4.
Soln.:
Note that y3 (x) − y1 (x) = x3 − x2 and y2 (x) − y3 (x) = 2x2 are solutions of the corresponding
homogeneous ODE. Therefore, we can take two linearly independent solutions of the corresponding homogeneous ODE as x2 and x3 .
...(1 Mark)
The general solution of the given non-homogeneous ODE can be taken as
Ax2 + Bx3 + x(x + 1) = c1 x2 + c2 x3 + x,
where c1 and c2 are arbitrary constants.
...(0.5 Mark)
Putting the initial condition, y(1) = 1 =⇒ c1 +c2 +1 = 2 and y 0 (1) = 4 =⇒ 2c1 +3c2 +1 = 4,
which gives c2 = 1, c1 = 0 and y(x) = x3 + x is the required solutions.
...(0.5 Mark)
Remark 1: If someone observes that y3 (x) itself satisfies the initial conditions, then he/she
should get full marks even though this would not work in general.
Remark 2: if students take y1 − y2 (x) and y1 (x) − y3 (x) as solutions of the homogeneous
ODE, they will get 1 mark. If they get the general solution of the non-homogeneous ODE
with A(y1 − y2 (x) + B(y1 (x) − y3 (x)) as the CF, they will get 0.5 mark in addition. They will
get the remaining 0.5 mark if they get the particular solution satisfying the initial condition
correctly.
Version: 1.246
Page 4 of 8
2013 Spring
QUIZ 1
MA 102 : ODE
[3pnts. ] 4. Find ALL the singular points of
(sin x)
y
d2 y dy
+ 2 = 0,
+
2
dx
dx x
and determine which of those are regular singular points. (Hint: you can assume that sin x
2
4
6
is analytic everywhere and the power series (1 − t3! + t5! − t7! + · · · )−1 has positive radius of
convergence.)
Soln.:
Rewriting the ODE in normalized form, we obtain
d2 y
dy
+ P (x) + Q(x)y = 0,
2
dx
dx
where P (x) =
1
sin x
and Q(x) =
1
.
x2 sin x
So the singular points are
x = nπ, n = 0; ±1, ±2, ±3, · · · .
...(0.5 Mark)
At x = 0, x2 Q(x) = sin1 x is not analytic (it is not even continuous at x = 0). Hence x = 0 is
not a regular singular point.
...(0.5 Mark)
At x = nπ, n = ±1, ±2, ±3, · · · , we have
x − nπ
sin(x − nπ)
t
= (−1)n
(t = x − nπ)
sin t
t
= (−1)n
t3
t5
tt
t − 3! + 5! − 7!
+ ···
−1
4
2
t
t
t6
n
= (−1) 1 − + − + · · ·
,
3! 5! 7!
(x − nπ)P (x) = (−1)n
so that (x − nπ)P (x) is analytic at x = nπ (t = 0).
...(1 Mark)
(−1)n (x − nπ)2
x2 sin(x − nπ)
(−1)n (x − nπ)
= (x − nπ)
(nπ + x − nπ)−1
sin(x − nπ)
i
1 h
x − nπ (x − nπ)2
= (x − nπ)P (x)
1−
+
+
·
·
·
nπ
nπ
(nπ)2
(x − nπ)2 Q(x) =
so that (x − nπ)2 Q(x) is analytic at x = nπ. Hence x = nπ, n = ±1, ±2, · · · are regular
singular points
. . .(1 Mark)
Version: 1.246
Page 5 of 8
2013 Spring
QUIZ 1
MA 102 : ODE
ADDITIONAL SPACE
Version: 1.246
Page 6 of 8
2013 Spring
QUIZ 1
MA 102 : ODE
ADDITIONAL SPACE
Version: 1.246
Page 7 of 8
2013 Spring
QUIZ 1
MA 102 : ODE
EXTRA SPACE FOR ROUGH WORK
Version: 1.246
Page 8 of 8