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Joint Entrance Examination
Mathematics
Differential Equations
Question 291
Question 291 of 490
Medium
MCQ
If
(
2
+
sin
x
)
d
y
d
x
+
(
y
+
1
)
cos
x
=
0
(2+\sin x)\dfrac{dy}{dx}+(y+1)\cos x=0
(
2
+
sin
x
)
d
x
d
y
+
(
y
+
1
)
cos
x
=
0
,
y
(
0
)
=
1
y(0)=1
y
(
0
)
=
1
, then
y
(
π
2
)
y\left(\dfrac{\pi}{2}\right)
y
(
2
π
)
is:
A
4
3
\dfrac43
3
4
B
1
3
\dfrac13
3
1
C
−
2
3
-\dfrac23
−
3
2
D
−
1
3
-\dfrac13
−
3
1
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Official solution
d
[
(
2
+
sin
x
)
(
y
+
1
)
]
=
0
⟹
(
2
+
sin
x
)
(
y
+
1
)
=
C
d[(2+\sin x)(y+1)]=0 \implies (2+\sin x)(y+1)=C
d
[(
2
+
sin
x
)
(
y
+
1
)]
=
0
⟹
(
2
+
sin
x
)
(
y
+
1
)
=
C
. Using
(
0
,
1
)
(0,1)
(
0
,
1
)
:
C
=
4
C=4
C
=
4
. At
x
=
π
/
2
x=\pi/2
x
=
π
/2
:
3
(
y
+
1
)
=
4
⟹
y
=
1
3
3(y+1)=4 \implies y=\dfrac13
3
(
y
+
1
)
=
4
⟹
y
=
3
1
.
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