Question

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

• A. 1
• B. 2
• C. 3
• D. none of these.

Answer 1

Answer: (A)

1

Answer 2

Putting $x = tan \,A$, $y = tan\, B$, we get $sec\, A + sec \,B = \lambda (tan\, A \,sec\, B - tan\, B\, sec\, A)$ $\Rightarrow \frac{cosA + cosB}{cosA \,cosB} = \lambda \left(\frac{sinA}{cosA \,cosB}-\frac{sinB}{cosA \,cosB}\right)$ $\Rightarrow cos\, A + cos \,B = \lambda \left(sin \,A - sin \,B\right)$ $\Rightarrow 2\,cos \frac{A+B}{2} cos \frac{A-B}{2}$ = $\lambda.2 cos\frac {A+B}{2}sin \frac {A-B}{2}$ $\Rightarrow tan \frac{A-B}{2} = \frac{1}{\lambda}$ $\Rightarrow A -B = 2\,tan^{-1}\left(\frac{1}{\lambda}\right) =$ constant $\Rightarrow tan^{-1} \,x - tan^{-1} \,y =$ constant $\Rightarrow \frac{1}{1+x^{2}}- \frac{1}{1+y^{2}} \frac{dy}{dx} = 0$ $\Rightarrow \frac{dy}{dx} = \frac{1+y^{2}}{1+x^{2}}$This is of degree $1$.