The degree of the differential equation satisfying the relation √ 1 + x 2 + √ 1 + y 2 = λ ( x √ 1 + y 2 − y √ 1 + x 2 ) is Options: A . &nbsp 1 B . &nbsp 2 C . &nbsp 3 D . &nbsp none of these

Answer: Option A : A On Putting x = t a n A , y = t a n B we get s e c A + s e c B = λ ( t a n A s e c B − t a n B s e c A ) c o s A + c o s B = λ ( s i n A − s i n B ) t a n ( A − B 2 ) = 1 λ t a n − 1 x − t a n − 1 y = 2 t a n − 1 1 λ On differentiating 1 1 + x 2 − 1 1 + y 2 d y d x = 0

The degree of the differential equation satisfying the relation √1+x2+&

Discussion Forum : Differential Equations

Question -

The degree of the differential equation satisfying the relation $\sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} = λ (x \sqrt{1 + y^{2}} - y \sqrt{1 + x^{2}})$ is

Options:

A . 1

B . 2

C . 3

D . none of these

Answer: Option A
:
A
On Putting

x = t a n A, y = t a n B

we get

s e c A + s e c B = λ (t a n A s e c B - t a n B s e c A) c o s A + c o s B = λ (s i n A - s i n B) t a n (\frac{A - B}{2}) = \frac{1}{λ} t a n^{- 1} x - t a n^{- 1} y = 2 t a n^{- 1} \frac{1}{λ}

On differentiating

\frac{1}{1 + x^{2}} - \frac{1}{1 + y^{2}} \frac{d y}{d x} = 0

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Discussion Forum : Differential Equations

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