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AIEEE 2006 Solved Paper
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© examsnet.com
Question : 137
Total: 143
The differential equation whose solution is
A
x
2
+
B
y
2
=
1
where
A
and
B
are arbitrary constants is of
second order and second degree
first order and second degree
first order and first degree
second order and first degree
Validate
Solution:
A
x
2
+
B
y
2
=
1
.....(1)
A
x
+
b
y
d
y
d
x
=
0
.....(2)
A
+
B
y
d
2
y
d
x
2
+
B
(
d
y
d
x
)
2
=
0
....(3)
From (2) and (3)
x
{
−
B
y
d
2
y
d
x
2
−
B
(
d
y
d
x
)
2
}
+
B
y
d
y
d
x
=
0
Dividing both sides by
−
B
, we get
⇒
x
y
d
2
y
d
x
2
+
x
(
d
y
d
x
)
2
−
y
d
y
d
x
=
0
Which is
D
E
of order 2 and degree 1 .
© examsnet.com
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