Particular solution of the differential equation $x(1+ y^2)dx-y(1+x^2)dy = 0$, given $y = 0$ when $x = 1$, is

$(1+x^2)=2(1+ y^2)$

The correct answer is Option (1) → $(1+x^2)=2(1+ y^2)$

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

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

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

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

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

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

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

$ \frac{1+x^{2}}{1+y^{2}} = k $

Apply condition $ y=0 $ when $ x=1 $:

$ \frac{1+1^{2}}{1+0^{2}} = k $

$ k = 2 $

$ \frac{1+x^{2}}{1+y^{2}} = 2 $

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

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