The general solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) is

(A)
\(\begin{array}{l}
\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0 \\
\therefore \quad\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=-\left(1+y^{2}\right) \Rightarrow \frac{d y}{d x}=\frac{-\left(1+y^{2}\right)}{x-e^{\tan ^{-1} y}} \\
\therefore \frac{d x}{d y}=\frac{\left(x-e^{\tan ^{-1} y}\right)}{-\left(1+y^{2}\right)} \Rightarrow \frac{d x}{d y}=\frac{-x}{\left(1+y^{2}\right)}+\frac{e^{\tan ^{-1} y}}{1+y^{2}} \\
\therefore \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{1+y^{2}} \\
\quad \text { I.F. }=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}
\end{array}\)
So, the general solution is
\(\begin{array}{l}
x \cdot e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}}{1+y^{2}} \cdot e^{\tan ^{-1} y} d x \\
\text { Put } e^{\tan ^{-1} y}=t \Rightarrow \frac{e^{\tan ^{-1} y}}{1+y^{2}} d y=d t \\
\therefore x \cdot e^{\tan ^{-1} y}=\int t d t \\
\therefore x \cdot e^{\tan ^{-1} y}=\frac{t^{2}}{2}+c \Rightarrow x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^{2}}{2}+c
\end{array}\)