The particular solution of \(\frac{y}{x} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2}\) when \(x=2, y=1\) is

\(\begin{aligned} & \frac{y}{x} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2} \\ & \Rightarrow \frac{y}{1+y^2} \mathrm{~d} y=\frac{x}{1+x^2} \mathrm{~d} x \\ & \Rightarrow \frac{1}{2} \int \frac{2 y \mathrm{~d} y}{1+y^2}=\frac{1}{2} \int \frac{2 x}{1+x^2} \mathrm{~d} x \\ & \Rightarrow \frac{1}{2} \log \left|1+y^2\right|=\frac{1}{2} \log \left|1+x^2\right|+\frac{1}{2} \log c \\ & \Rightarrow \log \frac{1+y^2}{1+x^2}=\log c \\ & \Rightarrow \frac{1+y^2}{1+x^2}=C\end{aligned}\)
Putting \(x=2, y=1\) we get \(C=\frac{2}{5}\)
\(
\begin{aligned}
& \Rightarrow \frac{1+y^2}{1+x^2}=\frac{2}{5} \\
& \Rightarrow 5\left(1+y^2\right)=2\left(1+x^2\right)
\end{aligned}
\)