Ans: (D)
Hint \(: x+y=z, \Rightarrow \frac{d y}{d x}=\frac{d z}{d x}-1\)
$$
\Rightarrow \frac{z^{2} d z}{a^{2}+z^{2}}=d x
$$
Integrating
$$
\begin{aligned}
&\Rightarrow x+y-a \tan ^{-1} \frac{x+y}{a}=x+c_{1} \Rightarrow \tan \left(\frac{y-c_{1}}{a}\right)=\frac{x+y}{a} \\
&\Rightarrow \tan \left(\frac{y+c}{a}\right)=\frac{x+y}{a}
\end{aligned}
$$
