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Test Index
JEE Main Math Class 12 Differential Equations Part 2 Questions
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© examsnet.com
Question : 16
Total: 124
The solution of the differential equation
(
x
2
+
y
2
)
dx
−
5
x
y
d
y
=
0
,
y
(
1
)
=
0
, is :
[9 Apr 2024 Shift 1]
|
x
2
−
4
y
2
|
5
=
x
2
|
x
2
−
2
y
2
|
6
=
x
|
x
2
−
4
y
2
|
6
=
x
|
x
2
−
2
y
2
|
5
=
x
2
Validate
Solution:
👈: Video Solution
(
x
2
+
y
2
)
dx
=
5
x
y
d
y
⇒
d
y
d
x
=
x
2
+
y
2
5
x
y
Put
y
=
V
x
⇒
V
+
x
dv
dx
=
1
+
V
2
5
V
⇒
xdv
dx
=
1
−
4
V
2
5
V
⇒
∫
V
1
−
4
V
2
dV
=
∫
dx
5
x
Let
1
−
4
V
2
=
t
⇒
−
8
V
dV
=
dt
⇒
∫
d
t
(
−
8
)
(
t
)
=
∫
d
x
5
x
⇒
−
1
8
ln
|
t
|
=
1
5
ln
|
x
|
+
ln
C
⇒
−
5
ln
|
t
|
=
8
ln
|
x
|
+
ln
K
⇒
ln
x
8
+
ln
|
t
5
|
+
ln
K
=
0
⇒
x
8
|
t
5
|
=
C
⇒
x
8
|
1
−
4
V
2
|
5
=
C
⇒
x
8
|
x
2
−
4
y
2
x
2
|
5
=
C
⇒
|
x
2
−
4
y
2
|
5
=
C
x
2
given
y
(
1
)
=
0
⇒
|
1
|
5
=
C
⇒
C
=
1
⇒
|
x
2
−
4
y
2
|
5
=
x
2
© examsnet.com
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