questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Differential Equations
Question 280
Question 280 of 490
Medium
MCQ
If
cos
x
d
y
d
x
−
y
sin
x
=
6
x
\cos x\dfrac{dy}{dx}-y\sin x=6x
cos
x
d
x
d
y
−
y
sin
x
=
6
x
,
0
<
x
<
π
/
2
0<x<\pi/2
0
<
x
<
π
/2
,
y
(
π
/
3
)
=
0
y(\pi/3)=0
y
(
π
/3
)
=
0
, then
y
(
π
/
6
)
y(\pi/6)
y
(
π
/6
)
equals:
A
π
2
2
3
\dfrac{\pi^2}{2\sqrt{3}}
2
3
π
2
B
−
π
2
2
-\dfrac{\pi^2}{2}
−
2
π
2
C
−
π
2
2
3
-\dfrac{\pi^2}{2\sqrt{3}}
−
2
3
π
2
D
−
π
2
4
3
-\dfrac{\pi^2}{4\sqrt{3}}
−
4
3
π
2
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
d
(
y
cos
x
)
=
6
x
d
x
⟹
y
cos
x
=
3
x
2
+
C
d(y\cos x)=6x\,dx \implies y\cos x=3x^2+C
d
(
y
cos
x
)
=
6
x
d
x
⟹
y
cos
x
=
3
x
2
+
C
. Using
(
π
/
3
,
0
)
(\pi/3,0)
(
π
/3
,
0
)
:
C
=
−
π
2
3
C=-\dfrac{\pi^2}{3}
C
=
−
3
π
2
. At
x
=
π
/
6
x=\pi/6
x
=
π
/6
:
y
⋅
3
2
=
π
2
12
−
π
2
3
=
−
π
2
4
⟹
y
=
−
π
2
2
3
y\cdot\dfrac{\sqrt{3}}{2} = \dfrac{\pi^2}{12}-\dfrac{\pi^2}{3} = -\dfrac{\pi^2}{4} \implies y=-\dfrac{\pi^2}{2\sqrt{3}}
y
⋅
2
3
=
12
π
2
−
3
π
2
=
−
4
π
2
⟹
y
=
−
2
3
π
2
.
Answer the question to unlock
Previous
Next