Birla Institute of Technology & Science, Pilani
MATH F211 (Mathematics III)
First Semester 2022-2023
Mid-Sem Examination (Closed Book)
Time: 90 Minutes Date: October 31, 2022 (Monday) Max. Marks: 105
1. Notations and symbols have their usual meaning.
2. Start new question on fresh page. Moreover, answer each subpart of a question in continuation.
3. Write END at the end of the last attempted question.
Q.1 (a) Examine the exactness of the differential equation
(y 4 + 2y)dx + (xy 3 + 2y 4 − 4x)dy = 0.
Hence, solve it by finding an appropriate integrating factor. [10]
(b) Find a general solution of the differential equation
2(x2 − 5x + 6)y ′′ + (2x − 3)y ′ − 8y = 0
in terms of Hypergeometric functions near x = 2. [16]
Q.2 Without using power series method:
(a) Find the general solution of the differential equation
√
4xy ′′ + 2y ′ + 9y = sin(3 x), x > 0.
[18]
(b) Find two linearly independent solutions of the differential equation
1 y
y ′′ − 1 + y ′ + = 0, x > 0,
x x
and hence, determine the solution satisfying the conditions y(1) = 0 and y ′ (1) = 1. [8]
Q.3 (a) Using the method of variation of parameters, find a particular solution of the differential equation
x2 y ′′ − 2x(1 + x)y ′ + 2(1 + x)y = x3 , x > 0.
[15]
(b) Using the Sturm-Comparison theorem, show that every non-trivial solution of the differential equation
2x2 y ′′ + 2x3 y ′ + [(1 + x2 )2 − (1 + x3 )]y = 0
has infinite number of zeros on the positive x-axis. [14]
Q.4 Consider the differential equation
x(x − 1)y ′′ + 3xy ′ + y = 0.
(a) Find all possible ordinary and singular points of the given differential equation and classify its singular points
(regular or irregular) with proper justification. [6]
(b) Find all possible Frobenius series solution(s) of the given differential equation near x = 0. [18]
***END***
Solution Key of Question - 2
√
2(a) The general solution of 4xy ′′ + 2y ′ + 9y = sin(3 x) ; (x > 0).
The reduced eqn can be transformed into an eqn of constant coeffs.
Q′ +2P Q
P = 1/(2x), Q = 9/(4x), Q3/2
= 0.
R√ √
Q dx = ( 32 ) √1
R
Let, z = x
dx = 3 x −→ [2 M]
y′ = 9 ′
2z yz
y ′′ = 81 ′′
y
4z 2 z
− 81 ′
y
4z 3 z
−→ [3 M]
4xy ′′ = 9yz′′ − z9 yz′
2y ′ = z9 yz′
√
Thus, 4xy ′′ + 2y ′ + 9y = sin(3 x) transformes into
yz′′ + yz = 1
9 sin z −→ [3 M]
√ √
Solution of the reduced eqn : y(x) = c1 sin(3 x) + c2 cos(3 x) −→ [2 M]
Particular solution : Let, yp (z) = z(A sin z + B cos z) −→ [2 M]
Putting into, yz′′ + yz′ = 1
9 sin z ⇒ A = 0, B = −1/18 −→ [4 M]
Finally, the general solution : √
√ √ x √
y(x) = c1 sin(3 x) + c2 cos(3 x) − 6 cos(3 x) −→ [2 M]
Alternate method for finding the particular solution : Method of variation
of parameters:
√ √
sin(3 x)
Writing 4xy ′′ + 2y ′ + 9y = sin(3 x) into the form y ′′ + 1 ′
2x y + 9
4x y = 4x
√ √
Two solutions√
of the reduced eqn : y1 = sin(3 x); y2 = cos(3 x)
R(x) = sin(3
4x
x)
; The Wronskian W (y1 , y2 ) = y1 y2′ − y2 y1′ = − 2√3 x
The particular solution : yp (x) = y1 v1 + y2 v2 −→ [2 M]
√
y2 R
dx = − cos(6 x)
R
Where, v1 = − W 36
√ √
R y1 R x sin(6 x)
and v2 = W dx = − 6 + 36 . −→ [4 M]
√ √ √
x cos(3 x) sin(3 x)
Thus, yp = y1 v1 + y2 v2 = − 6 + 36
Finally, the general solution : √
√ √ √
y(x) = c1 sin(3 x) + c2 cos(3 x) − 6x cos(3 x) −→ [2 M]
Note : For those who used the operator method for finding the particular
solution, marks are awarded accordingly.
(2b) Finding two LI solutions of the differential equation y ′′ − (1 + x1 )y ′ + xy =
0 ; (x > 0), and hence finding the solution satisfying y(1) = 0 and y ′ (1) = 1.
By inspection : y1 = ex is a solution. −→ [2 M]
Thus, y2 = v(x)y1 is another LI solution.
h R R (1+1/x)dx i R
Where, v(x) = e
e2x
dx = xe−x dx = −(1 + x)e−x
So, y2 = −(1 + x) −→ [4 M]
The general solution : y(x) = c1 ex + c2 (1 + x)
Putting y(1) = 0 and y ′ (1) = 1 ⇒ c1 = 2/e, c2 = −1
The particular solution : y(x) = 2e ex − (1 + x) −→ [2 M]
