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CBSE Questions for Class 12 Commerce Maths Differential Equations Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Maths
Differential Equations
Quiz 10
Solution of the differential equation
{
1
x
−
y
2
(
x
−
y
)
2
}
d
y
+
{
y
2
(
x
−
y
)
2
−
1
y
}
d
x
=
0
is
Report Question
0%
ln
|
x
y
|
+
x
y
x
−
y
=
c
0%
x
y
x
−
y
=
c
e
x
/
y
0%
ln
|
x
y
|
=
c
+
x
y
x
−
y
0%
n
o
n
e
o
f
t
h
e
s
e
The solution of the differential equation
x
2
d
y
d
x
cos
(
1
x
)
−
y
sin
(
1
x
)
=
−
1
;
w
h
e
r
e
y
→
−
1
a
s
x
→
∞
is
Report Question
0%
y
=
sin
(
1
x
)
−
cos
(
1
x
)
0%
y
=
x
+
1
x
sin
(
1
x
)
0%
y
=
cos
(
1
x
)
+
sin
(
1
x
)
0%
y
=
x
+
1
x
cos
(
1
x
)
Explanation
x
2
d
y
d
x
c
o
s
(
1
x
)
−
y
s
i
n
(
1
x
)
=
−
1
d
y
d
x
−
y
x
2
t
a
n
(
1
x
)
=
−
1
x
2
c
o
s
(
1
x
)
d
y
d
x
+
(
−
1
x
2
t
a
n
(
1
x
)
)
y
=
−
1
x
2
c
o
s
(
1
x
)
I
.
F
=
e
∫
−
1
x
2
t
a
n
(
1
x
)
d
x
1
x
=
t
−
d
x
x
2
=
d
t
=
e
∫
t
a
n
(
t
)
d
t
=
e
l
n
(
s
e
c
t
)
=
s
e
c
(
1
x
)
s
e
c
(
1
x
)
.
y
=
∫
1
x
2
c
o
s
(
1
x
)
.
s
e
c
(
1
x
)
=
−
∫
s
e
c
2
(
1
x
)
x
2
Let
1
x
=
t
−
d
x
x
2
=
d
t
=
∫
s
e
c
2
(
t
)
d
t
s
e
c
(
1
x
)
.
y
=
t
a
n
(
1
x
)
+
C
1.
−
1
=
0
+
C
C
=
−
1
s
e
c
(
1
x
)
.
y
=
t
a
n
(
1
x
)
−
1
,
y
=
s
i
n
(
1
x
)
−
c
o
s
(
1
x
)
The solution of the differential equation
d
y
d
x
=
x
2
+
y
2
+
1
2
x
y
satisfying
y
(
1
)
=
1
, is
Report Question
0%
a hyperbola
0%
a circle
0%
y
2
=
x
(
1
+
x
)
−
10
0%
(
x
−
2
)
2
+
(
y
−
3
)
2
=
5
Explanation
d
y
d
x
=
x
2
+
y
2
+
1
2
x
y
2
y
d
y
d
x
=
x
+
y
2
x
+
1
x
2
y
d
y
d
x
−
y
2
x
=
x
+
1
x
y
2
=
u
=
2
y
d
y
d
x
=
d
u
d
x
d
u
d
x
−
u
x
=
x
+
1
x
d
u
d
x
+
P
u
=
Q
1
x
d
u
d
x
−
u
x
2
=
1
+
1
x
2
∫
d
(
u
x
)
=
∫
(
1
+
1
x
2
)
d
x
⇒
u
x
x
−
x
−
1
+
C
⇒
y
2
x
=
x
−
1
x
+
C
⇒
y
2
=
x
2
−
1
+
x
The general solution of differential equation
d
y
d
x
=
sin
3
x
cos
2
x
+
x
e
x
Report Question
0%
y
=
1
5
cos
5
x
+
1
3
c
s
c
3
x
+
(
x
+
1
)
e
x
+
c
0%
y
=
1
5
cos
5
x
−
1
3
c
s
c
3
x
+
(
x
−
1
)
e
x
+
c
0%
y
=
−
1
5
cos
5
x
−
1
3
c
s
c
3
x
−
(
x
−
1
)
e
x
−
c
0%
None of these
The value of
lim
x
→
∞
y(x) obtained from the differential equation
d
y
d
x
=
y
−
y
2
, where y(0) = 2 is
Report Question
0%
1
0%
-1
0%
0
0%
2
2
−
e
Explanation
d
y
d
x
=
y
−
y
2
y
(
0
)
=
2
d
y
y
−
y
2
=
d
x
−
d
y
y
2
−
y
=
d
x
−
d
y
y
2
(
1
2
)
2
−
2
×
1
2
y
−
1
4
=
d
x
d
y
(
y
−
1
2
)
2
−
(
1
2
)
2
=
−
d
x
1
2
×
1
2
ln
|
y
−
1
2
−
1
2
y
−
1
2
+
1
2
|
=
−
x
+
c
ln
(
y
−
1
y
)
=
x
+
c
Given
y
(
0
)
=
2
ln
(
2
−
1
2
)
=
C
C
=
ln
(
1
2
)
ln
(
y
−
1
y
)
=
−
x
+
ln
(
1
2
)
ln
(
y
y
−
1
)
=
x
+
ln
(
2
)
y
y
−
1
=
e
x
+
ln
(
2
)
y
−
1
y
=
1
e
x
+
ln
(
2
)
1
−
1
y
=
e
−
(
x
+
ln
(
2
)
)
1
y
=
1
−
e
−
(
x
+
ln
(
2
)
]
y
=
1
1
−
e
−
[
x
+
ln
(
2
)
]
lim
x
→
∞
1
1
−
e
−
∞
=
1
1
−
0
=
1
The solution of the differential equation
d
y
d
x
+
1
x
tan
y
=
tan
y
sin
y
x
2
is
Report Question
0%
1
x
cos
y
=
1
2
x
2
+
K
0%
1
x
cot
y
=
1
2
x
2
+
K
0%
1
x
csc
y
=
1
2
x
2
+
K
0%
1
x
cos
x
=
1
2
x
2
+
K
The solution of the differential equation
(
3
x
y
+
y
2
)
d
x
+
(
x
2
+
x
y
)
d
y
=
0
is
Report Question
0%
x
2
(
2
x
+
y
)
=
3
0%
y
2
(
2
x
+
y
)
=
3
0%
x
2
y
(
2
x
+
y
)
=
3
0%
x
2
y
(
2
x
+
3
)
=
9
The solution of the differential equation
(
x
2
+
1
)
d
y
d
x
+
y
2
+
1
=
0
,
is
[
i
f
y
(
0
)
=
1
]
Report Question
0%
y
=
2
+
x
2
0%
y
=
(
1
+
x
)
(
1
−
x
)
0%
y
=
x
(
x
−
1
)
0%
y
=
(
1
−
x
)
(
1
+
x
)
The solution of differential equation
d
y
d
x
=
x
(
2
ln
x
+
1
)
sin
y
+
y
cos
y
is
Report Question
0%
y
sin
y
=
x
ln
x
+
c
0%
y
sin
y
=
x
2
ln
x
+
c
0%
sin
y
=
x
2
ln
x
+
c
0%
y
cos
y
=
x
2
ln
x
+
c
Explanation
Equation of family of circle is
(
n
+
a
)
2
+
(
y
−
a
)
2
=
a
2
or
x
2
+
y
2
+
2
a
x
−
2
a
y
+
a
2
=
0
------(1)
Diffentiating , we get
2
n
+
2
y
d
y
d
x
+
2
a
−
2
a
d
y
d
x
=
0
⇒
x
+
y
d
y
d
x
=
a
(
d
y
d
x
−
1
)
⇒
a
=
x
+
y
y
1
y
1
−
1
where
y
1
=
d
y
d
x
Substituting the value of a in eq
(
1
)
and Simplifying
⇒
(
x
y
1
−
x
+
x
+
y
y
1
)
+
(
y
y
1
−
y
−
x
−
y
y
1
)
2
=
(
x
+
(
y
y
1
)
)
2
⇒
(
x
+
y
)
2
[
(
y
1
)
2
+
1
]
=
(
x
+
y
y
1
)
2
The given differential equation is :
d
y
d
x
=
x
(
2
log
x
+
1
)
∼
y
+
y
cos
y
⇒
(
sin
y
+
y
cos
y
)
d
y
=
x
(
2
log
x
+
1
)
d
x
⇒
∫
(
sin
y
+
y
cos
y
)
d
y
=
∫
x
(
2
log
x
+
1
)
d
x
integrating both sides
⇒
−
cos
y
+
y
sin
y
−
∫
sin
y
d
y
=
2
log
x
x
2
2
−
2
∫
1
x
x
2
2
d
x
+
x
2
2
⇒
−
cos
y
+
y
sin
y
+
cos
y
=
x
2
log
x
−
x
2
2
+
x
2
2
+
c
⇒
y
sin
y
=
x
2
log
x
+
c
.......(2)
It is given that when
y
=
π
2
,
x
=
1
So, putting
y
=
π
2
,
x
=
1
in eq (2),
⇒
π
2
sin
π
2
=
1.
log
1
+
c
⇒
c
=
π
2
Putting
c
=
π
2
in eq (2) we oftain
y
sin
y
=
x
2
log
x
+
π
2
The equation of a curve passing through the point (0, 0) and whose differential equation is
y
′
=
e
x
sin
x
is
Report Question
0%
2
y
−
1
=
e
x
(
sin
x
−
cos
x
)
0%
2
y
−
1
=
e
−
x
(
cos
x
−
sin
x
)
0%
2
y
+
1
=
e
−
x
(
cos
x
−
sin
x
)
0%
N
o
n
e
o
f
t
h
e
s
e
Explanation
given,
d
y
d
x
=
e
x
sin
x
d
y
=
e
x
sin
x
.
d
x
∫
d
y
=
∫
e
x
sin
x
.
d
x
→
integrating both sides.
d
e
t
,
∫
e
x
sin
x
d
x
=
I
−
−
−
−
(
1
)
y
=
∫
e
x
↓
f
(
x
)
sin
x
↓
g
(
x
)
d
x
=
I
Since,
∫
f
(
x
)
g
(
x
)
d
x
=
f
(
x
)
.
∫
g
(
x
)
d
x
−
∫
[
f
′
(
x
)
∫
g
(
x
)
d
x
]
d
x
∴
⇒
e
x
.
∫
sin
x
d
x
−
∫
[
e
x
.
∫
sin
x
d
x
]
I
=
e
x
.
(
−
cos
x
)
−
∫
e
x
.
(
−
cos
x
)
.
d
x
I
=
−
e
x
cos
x
+
∫
e
x
↓
f
(
x
)
.
cos
x
↓
g
(
x
)
d
x
Now apply same formula as above
I
=
−
e
x
cos
x
.
+
[
e
x
.
∫
cos
x
.
d
x
−
∫
(
e
x
.
∫
cos
x
.
d
x
)
d
x
]
I
=
−
e
x
.
cos
x
+
[
e
x
.
sin
x
−
∫
e
x
.
sin
x
.
d
x
⏟
(
I
)
]
I
=
−
e
x
cos
x
+
e
x
sin
x
−
I
+
c
2
y
=
e
x
(
sin
x
−
cos
x
)
+
c
Since this curve passes through
(
0
,
0
)
therefore
2
×
0
=
e
0
(
sin
0
cos
0
)
+
c
c
=
1
∴
equation of curve
2
y
=
e
x
(
sin
x
−
cos
x
)
+
1
(
2
y
−
1
)
=
e
x
(
sin
x
−
cos
x
)
Solution of the differential equation
y
e
x
/
y
d
x
=
(
x
e
x
y
+
y
2
)
d
y
(
y
≠
0
)
is ?
Report Question
0%
e
x
/
y
=
x
+
C
0%
e
x
/
y
+
x
=
C
0%
e
y
/
x
=
x
+
C
0%
e
x
/
y
=
y
+
C
Solution of differential equation
x
d
y
=
(
y
−
x
2
−
y
2
)
d
x
is (where
c
is arbitrary constant)
Report Question
0%
y
=
x
cos
(
c
+
x
)
0%
y
=
x
tan
(
x
−
c
)
0%
y
=
x
csc
(
c
+
x
)
0%
none of these
Let
y
=
y
(
x
)
be the solution of the differential equation
(
1
−
x
2
)
d
y
d
x
−
x
y
=
1
,
x
∈
(
−
1
,
1
)
if
y
(
0
)
=
0
, then
y
(
1
2
)
is equal to
Report Question
0%
π
3
√
3
0%
π
√
3
0%
π
6
0%
π
3
If the general solution of the differential equation if
y
′
=
y
x
+
Φ
(
x
y
)
, for some function
Φ
, is given by
y
i
n
|
c
x
|
=
x
, where c is an arbitrary constant, then
Φ
(2) is equal to :
Report Question
0%
4
0%
1
4
0%
-4
0%
−
1
4
(
2
y
+
x
y
3
)
d
x
+
(
x
+
x
2
y
2
)
d
y
=
0
solution of differential equation is
Report Question
0%
3
x
2
y
+
x
3
y
2
=
c
0%
3
x
2
y
2
+
x
y
2
=
c
0%
y
x
2
+
x
y
3
=
c
0%
x
2
y
+
x
3
y
3
3
=
c
Solution of differential equation
x
6
d
y
+
3
x
5
y
d
x
=
x
d
y
−
2
y
d
x
is
Report Question
0%
x
3
y
=
y
x
2
+
C
0%
x
3
y
=
2
y
x
2
+
C
0%
x
3
y
2
=
y
x
2
+
C
0%
x
3
=
y
x
2
+
C
The solution of equation
d
y
d
x
=
1
x
+
y
+
1
Report Question
0%
x
=
c
e
y
−
y
−
2
0%
y
=
x
+
c
e
y
−
2
0%
x
+
c
e
y
−
2
−
2
=
0
0%
y
=
x
Let
y
=
y
(
x
)
be the solution of the differential equation
sin
x
d
y
d
x
+
y
cos
x
=
4
x
,
x
∈
(
0
,
π
)
, if
y
(
π
2
)
=
0
, then
y
(
π
6
)
is equal to
Report Question
0%
−
4
9
π
2
0%
4
9
√
2
π
2
0%
−
8
9
√
2
π
2
0%
−
8
9
π
2
Solution of the differential equation
d
r
=
a
(
r
sin
θ
d
θ
−
cos
θ
d
r
)
is
Report Question
0%
r
(
1
+
a
cos
θ
)
=
c
0%
r
(
1
+
a
cos
θ
)
=
a
c
0%
r
(
1
−
a
cos
θ
)
=
c
0%
r
(
1
−
a
cos
θ
)
=
a
c
The solutions of the differential equation
d
y
d
x
=
s
i
n
y
+
x
s
i
n
2
y
−
x
c
o
s
y
is
Report Question
0%
s
i
n
2
y
=
x
s
i
n
y
+
x
2
2
+
c
0%
s
i
n
2
y
=
x
s
i
n
y
−
x
2
2
+
c
0%
s
i
n
2
y
=
x
+
s
i
n
y
+
x
2
2
+
c
0%
s
i
n
2
y
=
x
−
s
i
n
y
+
x
2
2
+
c
The general solution of the differential equation
d
y
d
x
+
sin
x
+
y
2
=
sin
x
−
y
2
is
Report Question
0%
log
(
tan
y
2
)
=
c
−
2
sin
x
0%
log
(
tan
y
4
)
=
c
−
2
sin
(
x
2
)
0%
log
(
tan
(
y
2
+
π
4
)
)
=
c
−
2
sin
x
0%
None of the above
If
y
(
x
)
is the solution of the differential equation
(
x
+
2
)
d
y
d
x
=
x
2
+
4
x
−
9
,
x
≠
−
2
and
y
(
0
)
=
0
, then
y
(
−
4
)
is equal to
Report Question
0%
0
0%
1
0%
−
1
0%
2
The solution of differential equation,
(
x
+
tan
y
)
d
y
=
sin
2
y
d
x
is
Report Question
0%
x
c
o
t
y
=
1
2
log
|
c
o
s
e
c
2
y
−
c
o
t
2
y
|
+
c
0%
x
=
tan
y
+
c
√
tan
y
0%
x
=
cot
y
+
c
√
cot
y
0%
N
o
n
e
o
f
t
h
e
s
e
If a curve
y
=
f
(
x
)
passes through the point
(
1
,
−
1
)
and satisfies the differential equation,
y
(
1
+
x
y
)
d
x
=
x
d
y
, then
f
(
−
1
2
)
is equal to
Report Question
0%
−
4
5
0%
2
5
0%
4
5
0%
−
2
5
General solution of differential equation
x
2
(
x
+
y
d
y
d
x
)
+
(
x
d
y
d
x
−
y
)
√
x
2
+
y
2
=
0
is
Report Question
0%
1
√
x
2
+
y
2
+
y
x
=
c
0%
√
x
2
+
y
2
−
y
x
=
c
0%
√
x
2
+
y
2
+
y
x
=
c
0%
2
√
x
2
+
y
2
+
y
x
=
c
0%
1
√
x
2
+
y
2
−
y
x
=
c
Solution of the differential equation
d
y
d
x
=
y
(
1
+
x
)
x
(
y
−
1
)
is
Report Question
0%
log
|
x
y
|
+
x
−
y
=
C
0%
log
|
x
y
|
−
x
+
y
=
C
0%
log
|
x
y
|
+
x
+
y
=
C
0%
log
|
x
y
|
−
x
−
y
=
C
Solution of
(
x
+
y
−
1
)
d
x
+
(
2
x
+
2
y
−
3
)
d
y
=
0
is
Report Question
0%
y
+
x
+
l
o
g
(
x
+
y
−
2
)
=
c
0%
y
+
2
x
+
l
o
g
(
x
+
y
−
2
)
=
c
0%
2
y
+
x
+
l
o
g
(
x
+
y
−
2
)
=
c
0%
2
y
+
2
x
+
l
o
g
(
x
+
y
−
2
)
=
c
The solution of
x
d
y
x
2
+
y
2
=
(
y
x
2
+
y
2
−
1
)
d
x
is
Report Question
0%
y
=
x
cot
(
c
−
x
)
0%
cos
−
1
u
/
x
=
−
x
+
C
0%
y
=
x
tan
(
c
−
x
)
0%
y
2
/
x
2
=
x
tan
(
c
−
x
)
General solution of differential equation
x
2
d
y
+
y
(
x
+
y
)
d
x
=
0
is
Report Question
0%
2
x
+
y
=
A
x
2
Y
0%
2
x
−
y
=
A
x
2
y
0%
2
x
+
y
=
A
x
2
y
0%
x
−
2
y
=
A
x
2
y
If
y
(
t
)
is solution of
(
t
+
1
)
d
y
d
t
−
t
y
=
1
,
y
(
0
)
=
−
1
, then
y
(
1
)
=
Report Question
0%
1
4
0%
−
2
0%
−
1
2
0%
1
2
Solution of the differential equation
cos
x
d
y
=
y
(
sin
x
−
y
)
d
x
,
0
<
x
<
π
2
is
Report Question
0%
y sec x
=
tan x
+
c
0%
y sec x
=
sec x
+
c
0%
y sec x
=
(sec x
+
c) y
0%
sec x
=
y (tan x
+
c)
The solution of the differential equation
y
d
x
−
x
d
y
+
3
x
2
y
2
e
x
3
d
x
=
0
is
Report Question
0%
x
y
+
e
x
3
=
c
0%
x
y
−
e
x
=
c
0%
x
y
+
e
x
x
3
−
x
2
y
=
c
0%
x
y
−
e
x
3
=
c
Solution of the equation
(
x
+
y
)
2
d
y
d
x
=
4
,
y
(
0
)
=
0
is
Report Question
0%
y
=
2
t
a
n
−
1
(
x
+
y
2
)
0%
y
=
4
t
a
n
−
1
(
x
+
y
4
)
0%
y
=
4
t
a
n
−
1
(
x
+
y
2
)
0%
y
=
2
t
a
n
−
1
(
x
+
y
4
)
The solution of the differential equation
log
(
d
y
d
x
)
=
4
x
−
2
y
−
2
,
y
=
1
when
x
=
1
is
Report Question
0%
2
e
2
y
+
2
=
e
4
x
+
e
2
0%
2
e
2
y
−
2
=
e
4
x
+
e
2
0%
2
e
2
y
+
2
=
e
4
x
+
e
4
0%
3
e
2
y
+
2
=
e
3
x
+
e
2
the solution of
y
d
x
−
x
d
y
+
3
x
2
y
2
e
x
3
d
x
=
0
is
Report Question
0%
x
y
+
e
x
3
=C
0%
x
y
−
e
x
3
=0
0%
−
x
y
+
e
x
3
=C
0%
none of these
The solution of the differential equation
(
x
+
2
y
3
)
d
y
d
x
=
y
is
Report Question
0%
x
y
2
=
y
+
C
0%
x
y
=
y
2
+
C
0%
x
2
y
=
y
2
+
C
0%
x
y
=
x
2
+
C
General solution of the differential equation (x+y-2) dy=(x+y) dx is
Report Question
0%
y+x=log(y-x+1)+c
0%
y-x log (x+y-1)+c
0%
y-2x=log (x+y-1)+c
0%
y+2x=log (x+y+1)+c
The solution of
x
2
d
y
−
y
2
d
x
+
x
y
2
(
x
−
y
)
d
y
=
0
is?
Report Question
0%
l
n
|
x
−
y
x
y
|
=
y
2
2
+
c
0%
l
n
|
x
y
x
−
y
|
=
x
2
2
+
c
0%
l
n
|
x
−
y
x
y
|
=
x
2
2
+
c
0%
l
n
|
x
−
y
x
y
|
=
x
+
c
Solution of
(
1
+
x
y
)
y
d
x
+
(
1
−
x
y
)
x
d
y
=
0
is
Report Question
0%
log
x
y
+
1
x
y
=
c
0%
log
x
y
=
c
0%
log
x
y
−
1
x
y
=
c
0%
log
y
x
−
1
x
y
=
c
The general solution of the differential equation
y
−
x
d
y
d
x
=
y
2
cos
x
(
1
−
sin
x
)
is
Report Question
0%
y
(
sin
x
+
c
)
=
tan
x
+
sec
x
0%
y
(
sin
x
+
c
)
=
tan
x
−
sec
x
0%
x
y
=
sin
x
+
1
4
cos
2
x
+
c
0%
x
y
=
sin
x
+
1
8
cos
2
x
+
c
Solution of
d
y
d
x
+
2
x
y
=
y
is
Report Question
0%
y
=
c
e
x
−
x
2
0%
y
=
c
e
x
2
−
x
0%
y
=
c
e
x
0%
y
=
c
e
−
x
2
Let =y(x) be the solution of the differential equation, x
d
y
d
x
+
y
=
x
l
o
g
e
x
,
(
x
>
1
)
.
If
2
y
(
2
)
=
l
o
g
e
4
−
1
,
then y(e) is equal to :
Report Question
0%
−
e
2
2
0%
e
4
0%
e
2
4
0%
−
e
2
4
The solution of the Differential Equation
(
x
+
y
)
(
d
x
−
d
y
)
=
d
x
+
d
y
is
Report Question
0%
l
n
(
x
−
y
)
=
x
+
y
+
C
0%
l
n
(
x
+
y
)
=
x
−
y
+
C
0%
l
n
(
x
−
y
)
=
x
−
y
−
C
0%
n
o
n
e
o
f
t
h
e
s
e
The integrating factor of the equation
y
2
d
x
+
(
3
x
y
−
1
)
d
y
=
0
is
Report Question
0%
y
2
0%
y
3
0%
1
y
2
0%
1
y
3
The solution of the differential equation
d
y
d
x
=
sec
(
x
+
y
)
is
Report Question
0%
y
−
tan
x
+
y
2
=
c
0%
y
+
tan
x
+
y
2
=
c
0%
y
+
2
tan
x
+
y
2
=
c
0%
N
o
n
e
o
f
t
h
e
s
e
The solution of differential equation
d
y
/
d
x
−
y
/
x
+
tan
y
/
x
is:
Report Question
0%
x
=
C
′
cos
(
y
/
x
)
0%
x
=
C
′
sin
(
y
/
x
)
0%
x
=
C
′
csc
(
y
/
x
)
0%
n
o
n
e
o
f
t
h
e
s
e
The solution of the equation
d
y
d
x
=
(
1
+
x
)
y
(
y
−
1
)
x
is:
Report Question
0%
log
x
y
+
x
+
y
=
c
0%
log
(
x
y
)
+
x
−
y
=
c
0%
y
−
x
−
log
x
y
=
c
0%
N
o
n
e
o
f
t
h
e
s
e
Solution of
(
d
y
d
x
)
2
+
x
d
y
d
x
−
y
=
0
is
Report Question
0%
y
=
3
x
2
+
9
0%
y
=
3
x
+
9
0%
y
=
4
3
x
2
0%
y
=
9
x
+
3
Number of values of m
∈
N, for which
y
=
e
m
x
is a solution of the differential equation
D
3
y
−
3
D
2
y
−
4
D
y
+
12
y
=
0
, is?
Report Question
0%
0
0%
1
0%
2
0%
More than
2
The solution the differential equation
(
d
y
d
x
)
2
−
d
t
d
x
(
e
x
+
e
−
x
)
+
1
=
0
is/are
Report Question
0%
y
+
e
−
x
=
c
0%
y
−
e
−
x
=
c
0%
y
+
e
x
=
c
0%
y
−
e
x
=
c
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1
Not Answered
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Incorrect : 0
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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