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Byju's Answer
Standard XII
Mathematics
Derivative from First Principle
Solve the dif...
Question
Solve the different equation:-
(
tan
−
1
y
−
x
)
d
y
=
(
1
+
y
2
)
d
x
.
Open in App
Solution
(
tan
−
1
y
−
x
)
d
y
=
(
1
+
y
2
)
d
x
d
x
d
y
=
tan
−
1
y
−
x
1
+
y
2
d
x
d
y
=
tan
−
1
y
1
+
y
2
−
x
1
+
y
2
d
x
d
y
+
x
1
+
y
2
=
tan
−
1
y
1
+
y
2
Compare it with
d
x
d
y
+
p
x
=
q
p
=
1
1
+
y
2
q
=
tan
−
1
y
1
+
y
2
Integrating factor
=
e
∫
p
d
y
∫
p
d
y
=
∫
1
1
+
y
2
d
y
=
tan
−
1
y
I
.
F
=
e
tan
−
1
y
standard differential equation of this is
x
×
I
.
F
=
∫
q
×
I
.
F
+
C
x
×
e
tan
−
1
y
=
∫
tan
−
1
y
1
+
y
2
×
e
tan
−
1
y
d
y
+
c
...........(1)
∫
tan
−
1
y
1
+
y
2
e
tan
−
1
y
d
y
=
∫
tan
−
1
y
×
e
tan
−
1
y
×
(
1
1
+
y
2
d
y
)
put
tan
−
1
y
=
t
⇒
1
1
+
y
2
d
y
=
d
t
on substituting
=
∫
t
e
t
d
t
=
t
∫
e
t
d
t
−
∫
e
t
.1
d
t
=
t
e
t
−
e
t
=
e
t
(
t
−
1
)
put
t
=
tan
−
1
y
we get
=
e
tan
−
1
y
[
tan
−
1
y
−
1
]
put in equation (1)
x
tan
−
1
y
=
e
tan
−
1
y
[
tan
−
1
y
−
1
]
+
c
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Similar questions
Q.
Solve differential equation
(
1
+
y
2
)
d
x
=
(
tan
−
1
y
−
x
)
d
y
.
Q.
Solve the differential equation:
(
tan
−
1
y
−
x
)
d
y
=
(
1
+
y
2
)
d
x
.
Q.
Solve
(
1
+
y
2
)
d
x
=
(
tan
−
1
y
−
x
)
d
y
.
Q.
Solve :
(
1
+
y
2
)
d
x
=
(
t
a
n
−
1
y
−
x
)
d
y
.
Q.
Solution of the differential equation
(
1
+
y
2
)
d
x
=
(
1
−
t
a
n
−
1
y
−
x
)
d
y
is
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