QJEEM06Madd9e: JEE Main 2006 Mathematics - Class 12 JEEM06Madd9e Solution: The differential equation whose solution is where and are…

Question

Verified Solution • 2025–26 Session moderate

The differential equation whose solution is

A{x^2} + B{y^2} = 1

where

A

and

B

are arbitrary constants is of

Accepted Answer

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)

and

(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,

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

DE

of order

2

and degree

1.

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