The general solution of the differential equation $(1+x^2)dy-xdx=0$ is :

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

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

The correct answer is Option (1) → $\frac{e^{2y}}{1+x^2}=C$ $(1+x^2)dy-xdx=0$ $⇒\int dy=\int \frac{x}{1+x^2}dx$ $=\int dy = \frac{1}{2}\int\frac{2x}{1+x^2}dx$ $=y=\frac{1}{2}\log(1+x^2)+\log C$ $=2y=\log(1+x^2)+\log C$ $C(1+x^2)=e^{2y}⇒C=\frac{e^{2y}}{1+x^2}$