Question

The general solution of x(1 + y²)^1/2 dx + y(1 + x²)^1/2 dy = 0 is ______.

Options

• cos^–1 x + cos^–1 y = C

• x² + y² = (1 + x²)^1/2 + (1 + y²)^1/2 + C

• (1 + x²)^1/2 + (1 + y²)^1/2 = C

• tan^–1 x – tan^–1 y = C

Answer 1

The general solution of x(1 + y²)^1/2 dx + y(1 + x²)^1/2 dy = 0 is (1 + x²)^1/2 + (1 + y²)^1/2 = C.

Explanation:

x(1 + y²)^1/2 dx + y(1 + x²)^1/2 dy = 0

`\implies (xdx)/((1 + x^2)^(1//2)) + (ydy)/((1 + y^2)^(1//2)` = 0

Integrating we get

`2sqrt(1 + x^2) + 2sqrt(1 + y^2)` = 2c or

(1 + x²)^1/2 + (1 + y²)^1/2 = c