Transforms and Partial Differential Equations Solved MCQs

Multiple Choice Questions 44 Pages

Anna University

Mechanical Engineering

Contributed by

Arvind Ramson

MA8353 Transforms
and Partial
Differential
Equations
Mechanical
Engineering - Third
Semester
UNIT I PARTIAL
DIFFERENTIAL
EQUATIONS
TOPIC 1.1 FORMATION OF
PARTIAL DIFFERENTIAL
EQUATIONS - SINGULAR
INTEGRALS
1. Solve using the method of
separation of variables if u(x,0) = 10 e
-x
.

a) 10 e
-x
e
-t/3

b) 10 e
x
e
-t/3

c) 10 e
x/3
e
-t

d) 10 e
-x/3
e
-t

Answer: a

Explanation: u(x,t) = X(x) T(t)

Substituting in the given equation, X’T = 6
T’X + XT

which implies X = ce
(1+6k)x
which implies T = c’ e
kt
Therefore, u(x,t) = cc’ e
(1+6k)x
e
kt

Now, u(x,0) = 10 e
-x
= cc’e
(1+6k)x

Therefore, cc’ = 10 and k =
-1
⁄
3

Therefore, u(x,t) = 10 e
-x
e
-t/3
.
2. Find the solution of if
using the method of
separation of variables.

a)

b)

c)

d)

Answer: a

Explanation:

Substituting in the given equation,

which implies

which implies

Therefore k = -2 and cc’ =

Hence,
3. Solve the partial differential equation
using method of
separation of variables if

a)

b)
c)

d)

Answer: d

Explanation:

which implies

which implies

∂
u
∂
x
= 6
∂
u
∂
t
+
u
X
′
−
X
6
X
=
T
′
T
=
k
X
′
X
= 1 + 6
k
T
′
T
=
k
∂
u
∂
x
= 36
∂
u
∂
t
+ 10
u
∂
u
∂
x
(
t
= 0) = 3
e
−2
x
−3
2
e
−2
x
e
−
t
/3
3
e
x
e
−
t
/3
3
2
e
2
x
e
−
t
/3
3
e
−
x
e
−
t
/3
u
(
x
,
t
) =
X
(
x
)
T
(
t
)
X
′
T
= 36
T
′
X
+ 10
XT
X
′
X
=
k X
=
ce
kx
T
′
T
=
(
k
−10)
36
T
=
c
′
e
k
−10
36
t
∂
u
∂
x
(
t
= 0) = 3
e
−2
x
=
cc
′
ke
kx
−3
2
u
(
x
,
t
) =
−3
2
e
−2
x
e
−
t
/3
.
x
3 ∂
u
∂
x
+
y
2 ∂
u
∂
y
= 0
u
(0,
y
) = 10
e
5
y
.
10
e
5
2
x
2
e
5
y
10
e
−5
2
y
2
e
5
x
10
e
−5
2
y
2
e
−5
x
10
e
−5
2
x
2
e
5
y
u
(
x
,
t
) =
X
(
x
)
T
(
t
)
x
3
X
′
Y
+
y
2
Y
′
X
= 0
X
′
X
=
k
x
3
X
=
ce
k
2
x
2
Y
′
Y
=
−
k
y
2
Y
=
c
′
e
k
y
u
(
x
,
t
) =
cc
′
e
k
2
x
2
e
k
y
u
(0,
y
) = 10
e
5
y
=
cc
′
e
k
y
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Therefore k = 5 and cc’ = 10
Hence,
4. Solve the differential equation
using the method of
separation of variables if

a)

b)

c)

d)

Answer: a

Explanation:

which implies

which implies

Therefore

Hence cc’ = 9 and k = 17

Therefore,
5. Solve the differential equation
using the method of
separation of variables if .

a)

b)

c)

d)

Answer: c

Explanation:

k = 3 and cc’ = 1

Therefore
6. While solving a partial differential
equation using a variable separable method,
we assume that the function can be written as
the product of two functions which depend on
one variable only.
a) True
b) False
Answer: a
Explanation: If we have a function u(x,t),
then the function u depends on both x and t.
For using the variable separable method we
assume that it can be written as u(x,t) =
X(x).T(t) where X depends only on x and T
depends only on t.
7. While solving a partial differential
equation using a variable separable method,
we equate the ratio to a constant which?
a) can be positive or negative integer or zero
b) can be positive or negative rational number
or zero
c) must be a positive integer
d) must be a negative integer
Answer: b
Explanation: The constant can be any
rational number. For example, we use a
positive rational number to solve a 1-
Dimensional wave equation, we use a
negative rational number to solve 1-
Dimensional heat equation, 0 when we have
steady state. The choice of constant depends
on the nature of the given problem.
8. When solving a 1-Dimensional wave
equation using variable separable method, we
get the solution if _____________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything
Answer: b
Explanation: Since the given problem is 1-
Dimensional wave equation, the solution
should be periodic in nature. If k is a positive
number, then the solution comes out to be (c
7
e
px⁄c
+e
-px⁄c
c
8
)(c
7
e
pt
+e
-pt
c
8
) and if k is
positive the solution comes out to be
u
(
x
,
t
) = 10
e
−5
2
x
2
e
5
y
.
5
∂
u
∂
x
+ 3
∂
u
∂
y
= 2
u
u
(0,
y
) = 9
e
−5
y
.
9
e
17
5
x
e
−5
y
9
e
13
5
x
e
−5
y
9
e
−17
5
x
e
−5
y
9
e
−13
5
x
e
−5
y
u
(
x
,
t
) =
X
(
x
)
T
(
t
)
5
X
′
Y
+ 3
Y
′
X
= 2
XY
X
′
X
=
k
5
X
=
ce
k
5
x
Y
′
Y
= 2 −
k
/3
Y
=
c
′
e
2−
k
3
y
u
(
x
,
t
) =
cc
′
e
k
5
x
e
2−
k
3
y
u
(0,
y
) =
cc
′
=
e
2−
k
3
y
9
e
−5
y
u
(
x
,
t
) = 9
e
17
5
x
e
−5
y
.
x
2
∂
u
∂
x
+
y
2
∂
u
∂
y
=
u
u
(0,
y
) =
e
2
y
e
−3
y
e
2
x
e
3
y
e
2
x
e
−3
x
e
2
y
e
3
x
e
2
y
u
(
x
,
t
) =
X
(
x
)
T
(
t
)
x
2
X
′
Y
+
y
2
Y
′
X
=
XY
X
=
ce
−
k
x
Y
=
c
′
e
k
−1
y
u
(
x
,
t
) =
cc
′
e
−
k
x
e
k
−1
y
u
(0,
y
) =
e
2
y
=
cc
′
e
k
−1
y
u
(
x
,
t
) =
e
−3
x
e
2
y
.
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(ccos(px/c) + c’sin(px/c))(c’’cospt + c’’’sinpt).
Now, since it should be periodic, the solution
is (ccos(px/c) + c’sin(px/c))(c’’cospt +
c’’’sinpt).
9. When solving a 1-Dimensional heat
equation using a variable separable method,
we get the solution if ______________
a) k is positive
b) k is negative
c) k is 0
d) k can be anything
Answer: b
Explanation: Since this is a heat equation,
the solution must be a transient solution, that
is it should decay as time increases. This
happens only when k is negative and the
solution comes out to be (c’’cospx +
c’’’sinpx) (c e
-c
2
p
2
t
).
10. While solving any partial differentiation
equation using a variable separable method
which is of order 1 or 2, we use the formula
of fourier series to find the coefficients at last.
a) True
b) False
Answer: a
Explanation: After using the boundary
conditions, when we are left with only one
constant and one boundary condition, then we
use Fourier series coefficient formula to find
the constant.
TOPIC 1.2 SOLUTIONS OF
STANDARD TYPES OF FIRST
ORDER PARTIAL
DIFFERENTIAL EQUATIONS
1. First order partial differential equations
arise in the calculus of variations.

a) True

b) False

Answer: a
Explanation: The calculus of variations is a
type of analysis in the field of mathematics
(branch of calculus) which is used to find
maxima and minima of definite integrals.
2. The symbol used for partial derivatives, ∂,
was first used in mathematics by Marquis de
Condorcet.
a) True
b) False
Answer: a
Explanation: Partial derivatives are indicated
by the symbol ∂. This was first used in
mathematics by Marquis de Condorcet who
used it for partial differences.
3. What is the order of the equation,
?

a) Third Order

b) Second Order

c) First Order

d) Zero Order

Answer: c

Explanation: The equation having only first
derivative, i.e., are said to be first order
differential equation. Since the given equation
satisfies this condition, it is of first order.

4. In the equation, y= x
2
+c,c is known as the
parameter and x and y are known as the main
variables.

a) True

b) False

Answer: a

Explanation: Given: y= x
2
+c, where c is
known as an arbitrary constant. It is also
referred to as the parameter to differentiate it
from the main variables x and y.

5. Which of the following is one of the
criterions for linearity of an equation?

a) The dependent variable and its derivatives
should be of second order

xy
3
(
∂
y
∂
x
)
2
+
yx
2
+
∂
y
∂
x
= 0
∂
y
∂
x
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b) The dependent variable and its derivatives
should not be of same order
c) Each coefficient does not depend on the
independent variable
d) Each coefficient depends only on the
independent variable
Answer: d
Explanation: The two criterions for linearity
of an equation are:
The dependent variable y and its
derivatives are of first degree.
Each coefficient depends only on the
independent variable
6. Which of the following is a type of
Iterative method of solving non-linear
equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods
Answer: b
Explanation: There are 2 types of Iterative
methods, (i) Interpolation methods (or
Bracketing methods) and (ii) Extrapolation
methods (or Open-end methods).
7. Which of the following is an example for
first order linear partial differential equation?
a) Lagrange’s Partial Differential Equation
b) Clairaut’s Partial Differential Equation
c) One-dimensional Wave Equation
d) One-dimensional Heat Equation
Answer: a
Explanation: Equations of the form Pp + Qq
= R , where P, Q and R are functions of x, y,
z, are known as Lagrange’s linear equation.
8. What is the nature of Lagrange’s linear
partial differential equation?
a) First-order, Third-degree
b) Second-order, First-degree
c) First-order, Second-degree
d) First-order, First-degree
Answer: d
Explanation: Lagrange’s linear equation
contains only the first-order partial
derivatives which appear only with first
power; hence the equation is of first-order
and first-degree.
9. Find the general solution of the linear
partial differential equation, yzp+zxq=xy.
a) φ(x
2
-y
2
– z
2
)=0
b) φ(x
2
-y
2
, y
2
-z
2
)=0
c) φ(x
2
-y
2
, y
2
-x
2
)=0
d) φ(x
2
-z
2
, z
2
-x
2
)=0
Answer: b
Explanation: Given: yzp+zxq=xy
Here, the subsidiary equations are,

From the first two and last two terms, we get,
respectively,

or xdx-ydy=0, and
or ydx-zdy=0,

Integrating these we get two solutions

x
2
-y
2
=a , y
2
-z
2
=b

Hence, the general solution of the given
equation is,

φ(x
2
-y
2
, y
2
-z
2
)=0.

10. The equation 2 is an
example for Bernoulli’s equation.

a) False

b) True

Answer: b

Explanation: A first order, first degree
differential equation of the form,

is known as
Bernoulli’s equation.

11. A particular solution for an equation is
derived by eliminating arbitrary constants.

a) True

b) False

dx
yz
=
dy
zx
=
dz
xy
dx
yz
=
dy
zx
′
dx
zx
=
dy
xy
′
dy
dx
–
xy
=
y
−2
,
dy
dx
+
P
(
x
).
y
=
Q
(
x
).
y
a
,
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Answer: b
Explanation: A particular solution for an
equation is derived by substituting particular
values to the arbitrary constants in the
complete solution.
12. A partial differential equation is one in
which a dependent variable (say ‘y’) depends
on one or more independent variables (say
’x’, ’t’ etc.)
a) False
b) True
Answer: b
Explanation: A differential equation is
divided into two types, ordinary and partial
differential equations.
A partial differential equation is one in which
a dependent variable depends on one or more
independent variables.
Example:

TOPIC 1.3 LAGRANGE'S
LINEAR EQUATION
1. Singular solution for the Clairaut’s
equation is given by _______

a)

b) y
2
=-4ax

c) y
2
=4ax

d) x
2
=-2ay

Answer: c

Explanation: Let

Given equation is of the form y = px + f(p),
whose general solution is y = cx + f(c)

thus the general solution is
…………..(1), to find the value of c

we differentiate (1) partially w.r.t ‘c’ i.e

hence (1) becomes

y
2
=4ax is the singular solution.
2. Obtain the general solution for the equation
xp
2
+px-py+1-y=0 where p= .
a) y=cx+

b) x=cy-(c+1)

c) x=cy-

d) y=cx+(c+1)

Answer: a

Explanation: xp
2
+px-py+1-y=0

xp
2
+px+1=y(p+1)

……(1) thus
(1) is in the Clairaut’s equation form
y=px+f(p),

thus general solution is y=cx+ .
3. Find the general solution for the equation
(px-py)(py+x)=2p by reducing into Clairaut’s
form by using the substitution X=x
2
, Y=y
2
where p= .

a)

b)

c)

d)

Answer: b

Explanation:

now

now consider (px-py)(py+x)=2p substituting
the value of p we get

is in the Clairaut’s form

hence general solution is .
4. Find the general solution of the D.E 2y-
4xy’-log y’=0.

a)

b)

F
(
x
,
t
,
y
,
∂
y
∂
x
,
∂
y
∂
t
, … …) = 0
y
=
y
′
x
+
a
y
′
x
2
a
2
+
y
2
a
2
= 1
p
=
y
′
=
dy
dx
y
=
cx
+
a
c
0 =
x
−
a
c
2
→
c
2
=
a
x
→
c
=
√
a
x
y
=
√
a
x
x
+
a
√
x
a
→
y
= 2√
ax
dy
dx
1
c
+1
1
c
+1
y
=
xp
(
p
+1)+1
p
+1
or y
=
px
+
1
p
+1
1
c
+1
dy
dx
y
2
=
x
+
c
c
+1
y
2
=
cx
2
–
2
c
c
+1
x
2
=
cy
2
–
1
2
c
+1
x
2
=
y
2
+
c
2
c
+2
X
=
x
2
→
dX
dx
= 2
x
Y
=
y
2
→
dY
dy
= 2
y
p
=
dy
dx
=
dy
dY
dY
dX
dX
dx
and let P
=
dY
dx
p
=
1
2
y
∗
P
∗ 2
x or p
=
x
y
P i
.
e p
=
√
X
Y
P
(
√
X
Y
P
√
X
–
√
Y
)(
√
X
Y
P
√
Y
+
√
X
)
= 2
(
PX
−
Y
)
√
Y
(
P
+ 1)
√
X
= 2
√
X
Y
P
→ (
PX
−
Y
)
y
2
=
cx
2
–
2
c
c
+1
y
(
p
) =
2
c
p
– 1 +
logp
2
y
(
p
) =
c
2
p
– 2 +
logp
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c)

d)

Answer: a

Explanation: Let y’=p and hence given
equation is in Lagrange equation form

i.e 2y=4xp+log p …..(1) differentiating both
sides of the equation

2dy=4xdp+4pdx+ and dy=pdx

–> 2pdx=4xdp+4pdx+ –> -2pdx=4pdx+
-2p
(p≠0)…….this is a linear D.E for the function
x(p)

I.F is and solution is
x(p)

substituting back in (1) we
get

.
5. Find the general solution of the D.E y =
2xy’ – 3(y’)
2
.

a)

b)

c)

d)

Answer: b

Explanation: Let y’=p –> y = 2xp – 3p
2
….
(1) is in the Lagrange equation form

now differentiating we get dy=2xdp+2pdx-
6pdp and dy=pdx

thus pdx=2xdp+2pdx-6pdp→-pdx=2xdp-
6pdp –> …(2)

(2) is a linear D.E whose I.F=
hence its solution is

….substituting in (1) we get

Sanfoundry Global Education & Learning
Series – Ordinary Differential Equations.
TOPIC 1.4 LINEAR PARTIAL
DIFFERENTIAL EQUATIONS
OF SECOND AND HIGHER
ORDER WITH CONSTANT
COEFFICIENTS OF BOTH
HOMOGENEOUS AND NON-
HOMOGENEOUS TYPES
1. Solution of the differential equation
is _____________

a)

b)

c)

d)

Answer: a

Explanation:

–> x
2
ln x dy-xy dy=xy dx – y
2
ln y dx
…….dividing by x
2
y
2
then

on integrating we get

= c…. where c is a constant of
integration.
2. Solution of the differential equation
is ______

a)

b)

c)

d)

Answer: d

Explanation:

separating the variable
x
(
p
) =
−1
p
+
c
p
2
x
(
p
) =
1
2
p
+
c
p
1/2
dp
p
dp
p
dp
p
dx
dp
= 4
x
+
1
p
→
dx
dp
+
2
p
x
=
−1
2
p
2
e
∫
2
p
dp
=
e
logp
2
=
p
2
p
2
=
∫
p
2
∗
−1
2
p
2
dp
+
c
x
(
p
) =
−1
2
p
+
c
p
2
2
y
= 4
p
(
−1
2
p
+
c
p
2
)
+
logp
y
(
p
) =
2
c
p
– 1 +
logp
2
y
(
p
) =
p
1/2
+
c
2
p
y
(
p
) =
p
2
+
2
c
p
x
(
p
) = −
cp
+
c
p
2
x
(
p
) = 2
p
+
2
c
p
2
dx
dp
+
2
p
x
– 6 = 0
e
∫
2
p
dp
=
p
2
p
2
x
(
p
) =
∫
6
p
2
dp
+
c
→
x
(
p
) = 2
p
+
c
p
2
y
(
p
) = 2(2
p
+
c
p
2
)
p
− 3
p
2
=
p
2
+
2
c
p
.
dy
dx
=
y
(
x
−
ylny
)
x
(
xlnx
−
y
)
xlnx
+
ylny
xy
=
c
xlnx
−
ylny
xy
=
c
lnx
x
+
lny
y
=
c
lnx
x
–
lny
y
=
c
dy
dx
=
y
(
x
−
ylny
)
x
(
xlnx
−
y
)
lny
y
2
dy
–
1
xy
dy
=
1
xy
dx
–
lny
x
2
dx
(
lnx
(
1
−
y
2
dy
) +
1
xy
dx
) + (
lny
(
1
−
x
2
dx
) +
1
xy
d
d
(
lnx
y
) +
d
(
lny
x
) = 0
∫
d
(
lnx
y
) +
∫
d
(
lny
x
)
lnx
y
+
lny
x
dy
dx
=
e
3
x
−2
y
+
x
2
e
−2
y
e
2
y
3
=
e
3
x
3
+
x
2
2
+
c
e
3
y
(
e
2
x
+
x
3
)
6
+
c
e
2
y
(
e
3
x
+
x
3
)
6
+
c
e
2
y
2
=
e
3
x
3
+
x
3
3
+
c
dy
dx
=
e
3
x
−2
y
+
x
2
e
−2
y
dy
dx
=
e
−2
y
(
e
3
x
+
x
2
)
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e
2y
dy = (e
3x
+x
2
)dx…..integrating
∫ e
2y
dy =∫ (e
3x
+x
2
)dx
.
3. Solution of the differential equation sec
2
x
tan y dx + sec
2
y tan x dy=0 is _______

a) (sec x. sec y)=k

b) (sec x .tany)=k

c) (tan x. tany)=k

d) (sec x .tan x)+(sec y .tan y)=k

Answer: c

Explanation: sec
2
x tan y dx + sec
2
y tan x
dy=0

dividing throughout by tan y.tan x we get

……separating the
variable

now integrating we get

substituting tan x = t & tan y=p→sec
2
x
dx=dt & sec
2
y dy=dp

log t + log p = c –>log(tan x)+log(tan y) = c =
log k….since it is an unknown constant

log(tan x .tan y) = log k

(tan x tan y) = k is the solution.
4. Solution of the differential equation
is ______

a)

b)

c)

d) cot
-1
(4x+2y+1)=x+c

Answer: a

Explanation:

here we use substitution for

separating the variable and integrating

is the solution.
5. Solution of the differential equation
is ______

a) (y-x)-log(x(1+y))=c

b) log(x(1+y))=c

c) (y+x)-log(x)=c

d) (y-x)-log(y(1+x))=c

Answer: a

Explanation:

separating the variables & hence integrating

y – log(1+y) – log x – x = c

(y-x) – log(x(1+y)) = c is the solution.
6. Solve the differential equation
is _______

a) xp=(x
2
+2y
2
)
-3

b) x
2
p=(x
2
-2y
2
)
3

c) x
4
p=(x
2
-2y
2
)
-3

d) x
6
p=(x
2
+2y
2
)
3

Answer: b

Explanation: we can
clearly see that it is an homogeneous equation

hence substituting

separating the variables and integrating

…….substituting 1-
2v
2
=t→-4v dv=dt we get
e
2
y
2
=
e
3
x
3
+
x
3
3
+
c
sec
2
x
tanx
dx
+
sec
2
y
tany
dy
= 0
∫
sec
2
x
tanx
dx
+
∫
sec
2
y
tany
dy
=
c
→
∫
1
t
dt
+
∫
1
p
dp
=
c
dy
dx
= (4
x
+ 2
y
+ 1)
2
1
2
√
2
tan
−1
(
4
x
+2
y
+1
√
2
) =
x
+
c
1
√
2
cot
−1
(4
x
+ 2
y
+ 1) =
x
+
c
1
√
2
tan
−1
(
4
x
+2
y
+1
√
2
) =
c
dy
dx
= (4
x
+ 2
y
+ 1)
2
4
x
+ 2
y
+ 1 =
t
→ 4 + 2
dy
dx
=
dt
dx
→
dy
dx
=
1
2
dt
dx
– 2
1
2
dt
dx
– 2 =
t
2
dt
dx
= 2
t
2
+ 4
∫
1
2
t
2
+4
dt
=
∫
dx
1
2
√
2
tan
−1
t
√
2
=
x
+
c
1
2
√
2
tan
−1
(
4
x
+2
y
+1
√
2
) =
x
+
c
xy
dy
dx
= 1 +
x
+
y
+
xy
xy
dy
dx
= 1 +
x
+
y
+
xy
xy
dy
dx
= (1 +
x
) +
y
(1 +
x
) = (1 +
x
)(1 +
y
)
y
1+
y
dy
=
1+
x
x
dx
∫
y
1+
y
dy
=
∫
1+
x
x
x
∫
(1+
y
)−1
1+
y
dy
=
∫
1
x
dx
+
∫
1
dx
∫
1
dy
–
∫
1
1+
y
dy
–
log x
–
x
=
c
dy
dx
=
x
2
+
y
2
3
xy
dy
dx
=
x
2
+
y
2
3
xy
=
1+
y
2
x
2
3
y
x
y
=
vx
→
dy
dx
=
v
+
x
dv
dx
=
1+
v
2
3
v
x
dv
dx
=
1+
v
2
3
v
–
v
=
1−2
v
2
3
v
∫
3
v
1−2
v
2
dv
=
∫
1
x
dx
−3
4
log t
=
log x
+
log c
→
−3
4
log
(1 − 2
v
2
) =
−3
log
(
x
2
−2
y
2
x
2
) = 4
log cx
→
log
(
x
2
−2
y
2
x
2
)
−3
=
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is the
solution where p is constant.
7. The solution of differential equation
is ______

a)

b)

c)

d)

Answer: d

Explanation: we can
clearly see that it is an homogeneous equation

substituting

separating the variables and integrating we
get

log(sin v) = log x + log c

is the solution
where c is constant.
8. Particular solution of the differential
equation given y=-1 at x=1.

a) y=x

b) y+x=2

c) y=-x

d) y-x=2

Answer: c

Explanation:
……. is a homogeneous equation

thus put

separating the variables and integrating we
get

log(v
2
+1) – log(v+1) + log x = log c –>

where k=1/c

at x=1, y=-1 substituting we get 2k=0→k=0

thus the particular solution is y=-x.
UNIT II FOURIER SERIES
TOPIC 2.1 DIRICHLET'S
CONDITIONS
1. How many dirichlet’s conditions are there?

a) One

b) Two

c) Three

d) Four

Answer: c

Explanation: There are three dirichlet’s
conditions. These conditions are certain
conditions that a signal must possess for its
fourier series to converge at all points where
the signal is continuous.
2. What are the Dirichlet’s conditions?

a) Conditions required for fourier series to
diverge

b) Conditions required for fourier series to
converge if continuous

c) Conditions required for fourier series to
converge

d) Conditions required for fourier series to
diverge if continuous

Answer: b

Explanation: Dirichlet’s conditions are
Conditions required for fourier series to
converge. That is there are certain conditions
that a signal must posses for its fourier series
to converge at all points where the signal is
continuous.
x
6
(
x
2
−2
y
2
)
3
=
kx
4
→
x
2
p
= (
x
2
− 2
y
2
)
3
dy
dx
=
y
x
+
tan
y
x
cot
(
y
x
) =
xc
cos
(
y
x
) =
xc
sec
2
(
y
x
) =
xc
sin
(
y
x
) =
xc
dy
dx
=
y
x
+
tan
y
x
y
=
vx
→
dy
dx
=
v
+
x
dv
dx
=
v
+
tan v
∫
1
tanv
dv
=
∫
1
x
dx
sin v
=
xc
→
sin
(
y
x
) =
xc
dy
dx
=
y
2
−2
xy
−
x
2
y
2
+2
xy
−
x
2
dy
dx
=
y
2
−2
xy
−
x
2
y
2
+2
xy
−
x
2
=
y
2
x
2
–
2
y
x
–1
y
2
x
2
+
2
y
x
−1
y
=
vx
→
dy
dx
=
v
+
x
dv
dx
=
v
2
−2
v
−1
v
2
+2
v
−1
x
dv
dx
=
v
2
−2
v
−1
v
2
+2
v
−1
–
v
= −
(
v
3
+
v
2
+
v
+1)
v
2
+2
v
−1
= −
(
v
2
+1)(
v
+1)
v
2
+2
v
−1
∫
v
2
+2
v
−1
(
v
2
+1)(
v
+1)
dv
=
∫
−1
x
dx
∫
2
v
(
v
+1)−(
v
2
+1)
(
v
2
+1)(
v
+1)
dv
=
log c
–
log x
∫
(
2
v
v
2
+1
−
1
v
+1
)
dv
=
log c
–
log x
(
v
2
+1)
x
(
v
+1)
=
c
→
x
2
+
y
2
x
+
y
=
c
→
k
(
x
2
+
y
2
) = (
x
+
y
)
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3. What is the first Dirichlet’s condition?
a) Over any period, signal x(t) must be
integrable
b) Multiplication of the signals must be
continuous
c) x(t) should be continuous only
d) A signal can be integrable except break
points
Answer: a
Explanation: In the case of Dirichlet’s
conditions, the first property leads to the
integration of signal. It states that over any
period, signal x(t) must be integrable.
That is ∫|x(t)|dt<∞.
4. Is dirichlet’s condition possible in case of
discrete signals?
a) True
b) False
Answer: b
Explanation: Dirichlet’s conditions is not
possible in case of discrete signals. That is
these are certain conditions that a signal must
posses for its fourier series to converge at all
points where the signal is continuous only.
5. What guarantees that coefficient is finite in
a dirichlet’s condition?
a) First condition
b) Second condition
c) Third condition
d) Fourth condition
Answer: a
Explanation: The first property is:
That is ∫|x(t)|dt<∞
Now, X
n
= 1/T∫|x(t)e
-jwt
|dt ≤ 1/T∫|x(t)|dt
So, X
n
<∞.
6. What is the second dirichlet’s condition?
a) In any finite interval, x(t) is of bounded
variation
b) In most of a finite interval, x(t) is of
bounded variation
c) In any finite interval, x(t) is of unbounded
variation
d) In majority finite interval, x(t) is of
unbounded variation
Answer: a
Explanation: In any finite interval, x(t) is of
bounded variation. That is there are no more
than a finite number of maxima and minima
during a single period of the signal.
7. There are maxima and minima not possible
in dirichlet’s conditions.
a) True
b) False
Answer: b
Explanation: Maxima and minima are
possible if they are infinite number as stated
by the second dirichlet’s condition. In any
finite interval, x(t) is of bounded variation.
That is there are no more than a finite number
of maxima and minima during a single period
of the signal.
8. What is the third dirichlet’s condition?
a) Finite discontinuities in the infinite interval
b) Finite discontinuities in the finite interval
c) Infiinite discontinuities in the infinite
interval
d) Finite discontinuities in the all the intervals
Answer:b
Explanation: The third condition states that
in any finite interval of time, there is an only
a finite number of discontinuities. Hence,
finite discontinuities in the finite interval are
the correct option.
9. In the third condition, does each of the
discontinuities need to be finite?
a) All the time
b) Sometimes
c) Never
d) Rarely
Answer: a
Explanation: The third condition states that
in any finite interval of time, there is the only
finite number of discontinuities. And
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furthermore, each of these discontinuities
must be finite too.
10. What is the sum of the series at a point
when the signal is discontinuous?
a) Average of the previous limits
b) Previous limits are considered
c) Future limits are added
d) Average of left hand limit and right hand
limit are taken
Answer: d
Explanation: If the signal is discontinuous at
a point t, then an average of left hand limit
and right hand limit of the signal x(t) are
taken.
That is x(t) = ½[ x(t+)+x(t-)].
TOPIC 2.2 GENERAL FOURIER
SERIES Â€“ ODD AND EVEN
FUNCTIONS
1. Which of the following is not Dirichlet’s
condition for the Fourier series expansion?

a) f(x) is periodic, single valued, finite

b) f(x) has finite number of discontinuities in
only one period

c) f(x) has finite number of maxima and
minima

d) f(x) is a periodic, single valued, finite

Answer: d

Explanation: Dirichlet’s condition for
Fourier series expansion is f(x) should be
periodic, single valued and finite; f(x) should
have finite number of discontinuities in one
period and f(x) should have finite number of
maxima and minima in a period.
2. At the point of discontinuity, sum of the
series is equal to ___________

a)

b)

c)

d)

Answer: b
Explanation: When there is a point of
discontinuity, the value of the function at that
point is found by taking the average of the
limit of the function in the left hand side of
the discontinuous point and right hand side of
the discontinuous point. Hence the value of
the function at that point of discontinuity is
.
3. What is the Fourier series expansion of the
function f(x) in the interval (c, c+2π)?

a)

b)
c)

d)
Answer: a

Explanation: Fourier series expantion of the
function f(x) in the interval (c, c+2π) is given
by
where, a
0
is found by using n=0, in the
formula for finding a
n
. b
n
is found by using
sin(nx) instead of cos(nx) in the formula to
find a
n
.
4. If the function f(x) is even, then which of
the following is zero?

a) a
n

b) b
n

c) a
0

d) nothing is zero

Answer: b

Explanation: Since b
n
includes sin(nx) term
which is an odd function, odd times even
function is always odd. So, the integral gives
zero as the result.
5. If the function f(x) is odd, then which of
the only coefficient is present?

a) a
n

1
2
[
f
(
x
+ 0)–
f
(
x
− 0)]
1
2
[
f
(
x
+ 0) +
f
(
x
− 0)]
1
4
[
f
(
x
+ 0)–
f
(
x
− 0)]
1
4
[
f
(
x
+ 0) +
f
(
x
− 0)]
1
2
[
f
(
x
+ 0) +
f
(
x
− 0)]
a
0
2
+
∑
∞
n
=1
a
n
cos
(
nx
) +
∑
∞
n
=1
b
n
sin
(
nx
)
a
0
+
∑
∞
n
=1
a
n
cos
(
nx
) +
∑
∞
n
=1
b
n
sin
(
nx
)
a
0
2
+
∑
∞
n
=0
a
n
cos
(
nx
) +
∑
∞
n
=0
b
n
sin
(
nx
)
a
0
+
∑
∞
n
=0
a
n
cos
(
nx
) +
∑
∞
n
=0
b
n
sin
(
nx
)
a
0
2
+
∑
∞
n
=1
a
n
cos
(
nx
) +
∑
∞
n
=1
b
n
sin
(
nx
)
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Transforms and Partial Differential Equations Solved MCQs

Transforms and Partial Differential Equations Solved MCQs

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