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CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 11 - MCQExams.com
CBSE
Class 12 Commerce Maths
Differential Equations
Quiz 11
The solution of
x
2
d
y
−
y
2
d
x
+
x
y
2
(
x
−
y
)
d
y
=
0
, is?
Report Question
0%
l
o
g
|
x
−
y
x
y
|
=
y
2
2
+
C
0%
l
o
g
|
x
y
x
−
y
|
=
x
2
2
+
C
0%
log
|
x
−
y
x
y
|
=
x
2
2
+
C
0%
l
o
g
|
x
−
y
x
y
|
=
x
+
C
Solution of equation
d
y
d
x
=
y
d
d
(
ϕ
(
x
)
)
−
y
2
ϕ
(
x
)
is
Report Question
0%
y
=
ϕ
(
x
)
+
c
x
0%
y
=
ϕ
(
x
)
x
+
c
0%
y
=
ϕ
(
x
)
x
+
c
0%
y
=
ϕ
(
x
)
+
x
+
c
An equation containing an independent variable, dependent variable, and differential coefficients of the dependent variable with respect to the independent variable is called a/an _______
Report Question
0%
Differential equation
0%
Continuous function
0%
Discontinuous function
0%
Integral equation
The integrating factor of the equation
s
i
n
y
d
y
d
x
=
c
o
s
y
(
1
−
x
c
o
s
y
)
is
Report Question
0%
e
−
x
0%
e
x
0%
e
s
e
c
y
0%
N
o
n
e
o
f
t
h
e
s
e
Solution of differential equation
d
y
d
x
+
x
sin
2
y
=
sin
y
cos
y
is-
Report Question
0%
tan
y
=
(
x
−
1
)
+
C
e
−
x
0%
cot
y
=
(
x
−
1
)
+
C
e
−
x
0%
tan
y
=
(
x
−
1
)
e
x
+
c
0%
cot
y
=
(
x
−
1
)
e
x
+
c
Consider the differential equation
y
2
d
x
+
(
x
−
1
y
)
d
y
=
0
If
y
(
1
)
=
1
, then
x
is given by:
Report Question
0%
4
−
2
y
−
e
1
y
e
0%
3
−
1
y
+
e
1
y
e
0%
1
+
1
y
−
e
1
y
e
0%
1
−
1
y
+
e
1
y
e
Solution of differential equation
d
y
d
x
+
x
sin
2
y
=
sin
y
cos
y
is-
Report Question
0%
tan
y
=
(
x
−
1
)
+
c
e
−
x
0%
cot
y
=
(
x
−
1
)
+
c
e
−
x
0%
tan
y
=
(
x
−
1
)
e
x
+
C
0%
cot
y
=
(
x
−
1
)
e
x
+
C
(where C is an arbitrary constant)
Solve the differential equation:
(
e
y
+
1
)
cos
x
d
x
+
e
y
sin
x
d
y
=
0
Report Question
0%
sin
x
(
e
−
y
+
1
)
=
c
0%
sin
x
(
e
y
+
1
)
=
c
0%
sin
x
(
e
2
y
+
1
)
=
c
0%
cos
x
(
e
y
+
1
)
=
c
Let the population of rabbits surviving at a time t be governed by the differential equation
d
p
(
t
)
d
t
=
1
2
p
(
t
)
−
200.
If
p
(
0
)
=
100
, then p(t) equals:
Report Question
0%
600
−
500
e
t
/
2
0%
400
−
300
e
−
t
/
2
0%
400
−
300
e
t
/
2
0%
300
−
200
e
−
t
/
2
The solution of
y
5
x
+
y
−
x
d
y
d
x
=
0
is
Report Question
0%
x
4
4
+
1
5
(
x
y
)
5
=
C
0%
x
5
4
+
1
5
(
x
y
)
4
=
C
0%
(
x
y
)
5
+
x
4
4
=
C
0%
(
x
y
)
4
+
x
5
5
=
C
Suppose y=y(x) satisfies the differential equation
y
d
x
+
y
2
d
y
=
x
d
y
.
If
y
(
x
)
>
0
∀
x
ϵ
R
and y(1)=1, then y(-3) equals
Report Question
0%
1
0%
2
0%
3
0%
5
y
=
(
c
1
+
c
2
x
+
x
2
)
e
x
is a solution of
Report Question
0%
d
2
y
d
x
2
+
y
=
0
0%
d
2
y
d
x
2
+
x
y
=
0
0%
d
2
y
d
x
2
=
x
0%
d
2
y
d
x
2
−
2
d
y
d
x
+
y
=
2
e
x
An integrating factor of the differential equation
x
d
y
d
x
−
y
=
x
3
;
x
>
0
is ________
Report Question
0%
1
x
0%
-x
0%
−
1
x
0%
x
The solution of the differential equation
d
y
d
x
−
k
y
=
0
,
y
(
0
)
=
1
, approaches zero when
x
→
∞
,
if
Report Question
0%
k = 0
0%
k > 0
0%
k < 0
0%
k may be any real value
Solution of the differential equation
x
=
1
+
(
x
y
d
y
d
x
)
+
(
x
2
y
2
)
2
!
(
d
y
d
x
)
3
+
.
.
.
.
.
.
.
.
.
is
Report Question
0%
y
=
i
n
(
x
)
+
C
0%
y
=
(
i
n
x
)
2
+
C
0%
y
=
±
√
(
i
n
x
)
2
+
C
0%
x
y
=
x
y
+
K
Solution of differential equation
2
x
y
d
y
d
x
=
x
2
+
3
y
2
.
Report Question
0%
x
3
+
y
2
=
p
x
2
0%
x
2
2
+
y
3
x
=
y
2
+
p
0%
x
2
+
y
3
=
p
x
0%
x
2
+
y
2
=
p
x
3
y
=
3
√
2
x
Find
d
y
d
x
Report Question
0%
3
√
2
x
0%
2
√
2
x
0%
2
√
3
x
0%
2
√
4
x
Let
y
=
y
(
x
)
be a solution of the differential equation,
√
1
−
x
2
d
y
d
x
+
√
1
−
y
2
=
0
,
|
x
|
<
1
.
If
y
(
1
2
)
=
√
3
2
, then
y
(
−
1
√
2
)
is equal to:
Report Question
0%
√
3
2
0%
1
√
2
0%
−
√
3
2
0%
−
1
√
2
Explanation
√
1
−
x
2
d
y
d
x
+
√
1
−
y
2
=
0
⇒
d
y
√
1
−
y
2
+
d
x
√
1
−
x
62
=
0
Taking integration
⇒
∫
d
y
√
1
−
y
2
+
∫
d
x
√
1
−
x
2
=
0
⇒
sin
−
1
y
+
sin
−
1
x
=
c
....(1)
At
x
=
1
2
,
y
=
√
3
2
⇒
sin
−
1
(
√
3
2
)
+
sin
−
1
(
1
2
)
=
c
⇒
π
3
+
π
6
=
c
⇒
c
=
π
2
Put in equation (1)
sin
−
1
y
=
cos
−
1
x
y
(
−
1
√
2
)
=
sin
(
cos
−
1
(
−
1
√
2
)
)
=
1
√
2
∴
y
(
−
1
√
2
)
=
1
√
2
.
.
.
.
.
A
n
s
w
e
r
Hence option
′
B
′
is the answer.
Equation of the curve in which the subnormal is twice the square of the ordinate is given by
Report Question
0%
l
o
g
y
=
2
x
+
l
o
g
c
0%
y
=
c
e
2
x
0%
l
o
g
y
=
2
x
−
l
o
g
c
0%
None of these
The solution of
d
y
d
x
+
x
1
−
x
2
y
=
x
√
y
, is given by
Report Question
0%
3
√
y
+
(
1
−
x
2
)
=
C
(
1
−
x
2
)
1
/
4
0%
3
2
√
y
+
1
(
1
−
x
2
)
=
C
(
1
−
x
2
)
3
/
2
0%
3
√
y
−
(
1
−
x
2
)
=
C
(
1
−
x
2
)
3
/
2
0%
N
o
n
e
o
f
t
h
e
s
e
State true or false.
d
y
d
x
+
y
=
5
is a differential equation of the type
d
y
d
x
+
P
Y
=
Q
but it can be solved using the variable separable method also.
Report Question
0%
True
0%
False
The solution of
d
y
d
x
=
(
x
+
y
)
2
(
x
+
2
)
(
y
−
2
)
is given by
Report Question
0%
(
x
−
2
)
4
(
1
+
2
y
x
)
=
k
e
2
y
/
x
0%
(
x
−
2
)
4
(
1
+
2
(
y
−
2
)
(
x
+
2
)
)
=
k
e
2
(
y
−
2
)
(
x
+
2
)
0%
(
x
−
2
)
3
(
1
+
2
(
y
−
2
)
x
+
2
)
=
k
e
2
(
y
−
2
)
(
x
+
2
)
0%
n
o
n
e
o
f
t
h
e
s
e
If
(
y
2
−
2
x
2
y
)
d
x
+
(
2
k
y
2
−
x
3
)
d
y
=
0
then the value of
x
y
√
y
2
−
x
2
is
Report Question
0%
y
2
+
x
0%
x
y
2
0%
any constant
0%
None of these
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Incorrect : 0
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