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Joint Entrance Examination
Mathematics
Differential Equations
Question 339
Question 339 of 490
Medium
MCQ
The differential equation whose solution is
A
x
2
+
B
y
2
=
1
A{x^2} + B{y^2} = 1
A
x
2
+
B
y
2
=
1
where
A
A
A
and
B
B
B
are arbitrary constants is of
A
second order and second degree
B
first order and second degree
C
first order and first degree
D
second order and first degree
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Official solution
\[A{x^2} + B{y^2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\] \[Ax + by{{dy} \over {dx}} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\] \[A + By{{{d^2}y} \over {d{x^2}}} + B{\left( {{{dy} \over {dx}}} \right)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 3 \right)\] From
(
2
)
(2)
(
2
)
and
(
3
)
(3)
(
3
)
\[x\left\{ { - By{{{d^2}y} \over {d{x^2}}} - B{{\left( {{{dy} \over {dx}}} \right)}^2}} \right\} + By{{dy} \over {dx}} = 0\] Dividing both sides by
−
B
,
-B,
−
B
,
we get \[ \Rightarrow xy{{{d^2}y} \over {d{x^2}}} + x{\left( {{{dy} \over {dx}}} \right)^2} - y{{dy} \over {dx}} = 0\] Which is
D
E
DE
D
E
of order
2
2
2
and degree
1.
1.
1.
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