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  • Please,please help me - Ordinary Differential Equations - BITSAT

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)

best_answer

 

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

Posted by

prateek

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