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

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    None of these

 

Answers (1)

 

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable 
\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )

- wherein

eg:

  \frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0

 

 \sqrt{1+x^{2}}+\sqrt{1+y^{2}}=\lambda (x\sqrt{1+y^{2}}-y\sqrt{1+x^{2}}) is

Put x= \tan \alpha, y= \tan \beta

\therefore \sec \alpha + \sec \beta= \lambda (\tan \alpha \sec \beta - \tan \beta \sec \alpha)

\Rightarrow \cos \alpha + \cos \beta= \lambda (\sin \alpha -\sin \beta)

\Rightarrow \cot (\frac{\alpha + \beta}{2})= \lambda

\alpha + \beta = 2 \cot^{-1} \lambda=c

\therefore \tan ^{-1}x + \tan ^{-1}y = C

\therefore \frac {1}{1+x^{2}} +\frac {1}{1+y^{2}} \frac{dy}{dx}=0

\therefore \frac {dy}{dx}= - (\frac{1+y^{2}}{1+x^{2}})

 


Option 1)

1

Correct

Option 2)

2

Incorrect

Option 3)

3

Incorrect

Option 4)

None of these

Incorrect

Preparation Products

Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout BITSAT-JEE Main 2020

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 14999/- ₹ 7999/-
Buy Now
Exams
Articles
Questions