Differential EquationsmediumFree

Degree of DE from √(1+x²)+√(1+y²)=λ(...) | JEE

JEE Maths question with a full step-by-step solution.

Question

The degree of the differential equation satisfying the relation

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

is:

1

correct

2

3

DNone of these

Solution

Put

x = \tan A

and

y = \tan B

. Then

\sqrt{1+x^2} = \sec A

\sqrt{1+y^2} = \sec B

, and the relation becomes

\sec A + \sec B = \lambda(\tan A\sec B - \tan B\sec A).

Multiply through by

\cos A\cos B

\cos B + \cos A = \lambda(\sin A - \sin B).

Using sum-to-product,

\dfrac{\cos A + \cos B}{\sin A - \sin B} = \cot\dfrac{A-B}{2} = \lambda \Rightarrow A - B = 2\cot^{-1}\lambda

, i.e.

\tan^{-1}x - \tan^{-1}y = 2\cot^{-1}\lambda \ (\text{constant}).

Differentiate:

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

This contains

\dfrac{dy}{dx}

to the first power, so the degree is

1

. Correct answer: (1)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.