Question

The solution of differential equation `(x^2 + 1) "dy"/"dx" + (y^2 + 1) = 0` is ______.

विकल्प

• x + y = c

• (x² + 1)(y² + 1) = c

• x² = y² + c

• tan^-1 x + tan^-1 y = c

Answer 1

The solution of differential equation `(x^2 + 1) "dy"/"dx" + (y^2 + 1) = 0` is tan^-1 x + tan^-1 y = c.

Explanation:

We have, differential equation `(x^2 + 1) "dy"/"dx" + (y^2 + 1) = 0`

`=> "dx"/(x^2 + 1) = - "dy"/("y"^2 + 1)`

On the integrating both sides, we get

`int "dx"/(x^2 + 1) = - int "dy"/("y"^2 + 1)`

⇒ tan^-1 x = - tan^-1 y + c

⇒ tan^-1 x + tan^-1 y = c