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Byju's Answer
Standard XII
Mathematics
General Solutions of (sin theta)^2 = (sin alpha)^2 , (cos theta)^2 = (cos alpha)^2 , (tan theta)^2 = (tan alpha)^2
Find the part...
Question
Find the particular solution of the differential equation
d
x
d
y
+
x
cot
y
=
2
y
+
y
2
cot y, y ≠ 0 given that x = 0 when
y
=
π
2
.
Open in App
Solution
We
have
,
d
x
d
y
+
x
cot
y
=
2
y
+
y
2
cot
y
.
.
.
.
.
1
Clearly
,
it
is
a
linear
differential
equation
of
the
form
d
x
d
y
+
P
x
=
Q
where
P
=
cot
y
and
Q
=
2
y
+
y
2
cot
y
∴
I
.
F
.
=
e
∫
P
d
y
=
e
∫
cot
y
d
y
=
e
log
sin
y
=
sin
y
Multiplying
both
sides
of
1
by
I
.
F
.
=
sin
y
,
we
get
sin
y
d
x
d
y
+
x
cot
y
=
sin
y
y
2
cot
y
+
2
y
⇒
sin
y
d
x
d
y
+
x
cos
y
=
y
2
cos
y
+
2
y
sin
y
Integrating
both
sides
with
respect
to
y
,
we
get
x
sin
y
=
∫
y
2
I
cos
y
II
d
y
+
∫
2
y
sin
y
d
y
+
C
⇒
x
sin
y
=
y
2
∫
cos
y
d
y
-
∫
d
d
y
y
2
∫
cos
y
d
y
d
y
+
∫
2
y
sin
y
d
y
+
C
⇒
x
sin
y
=
y
2
sin
y
-
∫
2
y
sin
y
d
y
+
∫
2
y
sin
y
d
y
+
C
⇒
x
sin
y
=
y
2
sin
y
+
C
Now
,
y
=
π
2
at
x
=
0
∴
0
×
sin
π
2
=
π
4
2
sin
π
2
+
C
⇒
C
=
-
π
4
2
Putting
the
value
of
C
,
we
get
x
sin
y
=
y
2
sin
y
-
π
4
2
Hence
,
x
sin
y
=
y
2
sin
y
-
π
4
2
is
the
required
solution
.
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