Question

Find the general solution of differential equation:

x√(1 + y²) dx + y√(1 + x²) dy = 0

Answer 1

\(x \sqrt {1 + y^2} dx + y \sqrt{1 + x^2}dy = 0\)

⇒ \(x \sqrt{1+ y^2} dx = -dy (y\sqrt{1 + x^2})\)

⇒ \(\frac{dy}{dx} = \frac{-x\sqrt{1 + y^2}}{y \sqrt{1 + x^2}}\)

⇒ \(\frac y{\sqrt{1 + y^2}}dy = \frac{-x}{\sqrt{1 + x^2}} dx\)

⇒ \( \int \frac x{\sqrt{1 + x^2}}+ \int \frac y{\sqrt{1 + y^2}} = C\)

⇒ \( \frac 12 \int \frac{2x}{\sqrt{1 + x^2}} + \frac 12 \int \frac{2y}{\sqrt{1 + y^2}} dy = C\)

⇒ \(\sqrt{1 + x^2} + \sqrt{1 + y^2} = C\)

Answer 2

It is given that