The general solution of the differential equation $x(1 + y^2)dx + y(1 + x^2)dy = 0$ is
Answer & explanation
Correct answer: option 1
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:
$(1+x^{2})(1+y^{2}) = C$