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

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

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

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