1
You visited us
1
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard IX
Mathematics
Applications of Pair of Equations
x 2+y 2 d y d...
Question
x
2
+
y
2
d
y
d
x
=
8
x
2
-
3
x
y
+
2
y
2
Open in App
Solution
We
have
,
x
2
+
y
2
d
y
d
x
=
8
x
2
-
3
x
y
+
2
y
2
⇒
d
y
d
x
=
8
x
2
-
3
x
y
+
2
y
2
x
2
+
y
2
This
is
a
homogeneous
differential
equation
.
Putting
y
=
v
x
and
d
y
d
x
=
v
+
x
d
v
d
x
,
we
get
v
+
x
d
v
d
x
=
8
x
2
-
3
v
x
2
+
2
v
2
x
2
x
2
+
v
2
x
2
⇒
x
d
v
d
x
=
8
-
3
v
+
2
v
2
1
+
v
2
-
v
⇒
x
d
v
d
x
=
8
-
4
v
+
2
v
2
-
v
3
1
+
v
2
⇒
x
d
v
d
x
=
4
2
-
v
+
v
2
2
-
v
1
+
v
2
⇒
x
d
v
d
x
=
4
+
v
2
2
-
v
1
+
v
2
⇒
1
+
v
2
4
+
v
2
2
-
v
d
v
=
1
x
d
x
Integrating
both
sides
,
we
get
∫
1
+
v
2
4
+
v
2
2
-
v
d
v
=
∫
1
x
d
x
.
.
.
.
.
(
1
)
Let
us
consider
the
left
hand
side
of
(
1
)
.
Using
partial
fraction
,
Let
1
+
v
2
4
+
v
2
2
-
v
=
A
v
+
B
4
+
v
2
+
C
2
-
v
⇒
1
+
v
2
=
A
v
2
-
v
+
B
2
-
v
+
C
4
+
v
2
⇒
1
+
v
2
=
2
A
v
-
A
v
2
+
2
B
-
B
v
+
4
C
+
C
v
2
Comparing
the
coefficients
of
both
sides
,
we
get
2
A
-
B
=
0
-
A
+
C
=
1
&
2
B
+
4
C
=
1
Solving
these
three
equations
,
we
get
A
=
-
3
8
,
B
=
-
3
4
and
C
=
5
8
∴
1
+
v
2
4
+
v
2
2
-
v
=
-
3
8
v
-
3
4
4
+
v
2
+
5
8
2
-
v
.
.
.
.
.
(
2
)
From
(
1
)
and
(
2
)
,
we
get
∫
-
3
8
v
-
3
4
4
+
v
2
+
5
8
2
-
v
=
∫
1
x
d
x
⇒
-
3
8
∫
v
v
2
+
4
d
v
-
3
4
∫
1
v
2
+
4
d
v
+
5
8
∫
1
2
-
v
d
v
=
∫
1
x
d
x
⇒
-
3
16
log
v
2
+
4
-
3
4
×
2
tan
-
1
v
2
-
5
8
log
2
-
v
=
log
x
+
log
C
⇒
-
3
4
×
2
tan
-
1
v
2
=
log
C
x
2
-
v
5
8
v
2
+
4
3
16
⇒
e
-
3
8
tan
-
1
v
2
=
C
x
2
-
v
5
8
v
2
+
4
3
16
Putting
v
=
y
x
,
we
get
⇒
e
-
3
8
tan
-
1
y
2
x
=
C
x
2
-
y
x
5
8
y
x
2
2
+
4
3
16
⇒
e
-
3
8
tan
-
1
y
2
x
=
C
x
×
1
x
2
x
-
y
5
8
y
2
+
4
x
2
3
16
⇒
e
-
3
8
tan
-
1
y
2
x
=
C
2
x
-
y
5
8
y
2
+
4
x
2
3
16
Hence
,
e
-
3
8
tan
-
1
y
2
x
=
C
2
x
-
y
5
8
y
2
+
4
x
2
3
16
is
the
required
solution
.
Suggest Corrections
0
Similar questions
Q.
The pairs of straight lines
x
2
−
3
x
y
+
2
y
2
=
0
and
x
2
−
3
x
y
+
2
y
2
+
x
−
2
=
0
form a
Q.
Find the HCF of the following pair of polynomials:
x
2
−
x
y
−
2
y
2
and
x
2
+
3
x
y
+
2
y
2
Q.
Find the GCD of the following
x
2
+
3
x
y
+
2
y
2
,
x
2
+
5
x
y
+
6
y
2
Q.
If the hyperbolas,
x
2
+
3
x
y
+
2
y
2
+
2
x
+
3
y
+
2
=
0
and
x
2
+
3
x
y
+
2
y
2
+
2
x
+
3
y
+
c
=
0
are conjugate of each other, the value of
c
is equal to
Q.
y
1
+
x
2
+
x
1
+
y
2
d
y
d
x
=
0
View More
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Related Videos
MATHEMATICS
Watch in App
Explore more
Applications of Pair of Equations
Standard IX Mathematics
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
AI Tutor
Textbooks
Question Papers
Install app