The general solution of the differential equation $x(1 + y^2)dx + y(1 + x^2)dy = 0$ is

$(1+ y^2) (1+x^2)= C$: where C is an arbitrary constant

The correct answer is Option (1) → $(1+ y^2) (1+x^2)= C$: where C is an arbitrary constant

Given differential equation:

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

Rewrite:

$x(1+y^{2})\,dx = -\,y(1+x^{2})\,dy$

Divide both sides by $(1+x^{2})(1+y^{2})$:

$\frac{x}{1+x^{2}}\,dx = -\,\frac{y}{1+y^{2}}\,dy$

Integrate both sides:

$\int \frac{x}{1+x^{2}}\,dx = -\int \frac{y}{1+y^{2}}\,dy$

$\frac{1}{2}\log(1+x^{2}) = -\frac{1}{2}\log(1+y^{2}) + C$

Multiply by 2:

$\log(1+x^{2}) + \log(1+y^{2}) = C$

Combine logs:

$\log\!\big((1+x^{2})(1+y^{2})\big)=C$

Hence the general solution is:

$(1+x^{2})(1+y^{2}) = C$