The equation of the curve passing through (1, 0) and Satisfying the differential equation

(1 + y²) dx - xydy = 0 is -

Given a differential equation,

(1 + y2) dx - xydy = 0

or, (1 + y2) dx = xydy

It can be written as,

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

Integrating both sides,

ln (x) = \(\frac{1}{2}ln\ (1+y^2)+ln \ C\)

or, 2 ln (x) = ln (1 + y²) + 2 ln (C)

or, ln (x²) = ln [(1 + y2) (C²)]

Let, C² = K

or, x² = K (1 + y²) .... (1)

Since, the curve passing through (1, 0),

1 = K (1 + 0)

or, K = 1

From equation (1),

x2 = (1 + y2)

Hence,

x2 - y2 = 1