Let y=y(x), y>0, be a solution curve of the differential equation $(1+{\mathrm{x}}^2)\mathrm{dy}=\mathrm{y}(\mathrm{x}-\mathrm{y})\mathrm{dx}.$ If y(0)=1 and $\mathrm{y}\left(2\sqrt{2}\right)=\mathrm{\beta },$ then

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}(\mathrm{x}-\mathrm{y})}{1+{\mathrm{x}}^2}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{xy}}{1+{\mathrm{x}}^2}-\frac{{\mathrm{y}}^2}{1+{\mathrm{x}}^2}$
Divided by y²
$\Rightarrow \frac{1}{{\mathrm{y}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{x}}{1+{\mathrm{x}}^2}\left(\frac{1}{\mathrm{y}}\right)=\frac{-1}{1+{\mathrm{x}}^2}$
Put $\frac{-1}{\mathrm{y}}=\mathrm{t}$ then, $\frac{1}{{\mathrm{y}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$
$\therefore \frac{\mathrm{dt}}{\mathrm{dx}}+\left(\frac{\mathrm{x}}{1+{\mathrm{x}}^2}\right)\mathrm{t}=\frac{-1}{1+{\mathrm{x}}^2}$   (linear form)
I.F ${\mathrm{e}}^{\frac{1}{2}\int \frac{2x}{1+{\mathrm{x}}^2}\mathrm{dx}}=\sqrt{1+{\mathrm{x}}^2}$
$\therefore$Given solution is $\mathrm{t}\sqrt{1+{\mathrm{x}}^2}=\int \frac{-1}{1+{\mathrm{x}}^2}\sqrt{1+{\mathrm{x}}^2}\mathrm{dx}$
$\Rightarrow \frac{-\sqrt{1+{\mathrm{x}}^2}}{\mathrm{y}}=-\int \frac{1}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}$
$\Rightarrow \frac{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{y}}=\ln |\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}|+\mathrm{c}$
$$\begin{gathered} \mathrm{y}\left(0\right)=1\Rightarrow 1=\ln 1+\mathrm{c}\Rightarrow \mathrm{c}=1 \\ \therefore \frac{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{y}}=\ln |\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}|+1 \\ \mathrm{\beta }=\mathrm{y}\left(2\sqrt{2}\right) \\ \Rightarrow \frac{3}{\mathrm{\beta }}=\ln |2\sqrt{2}+3|+\ln \mathrm{e}=\ln \mathrm{e}(3+2\sqrt{2}) \\ \Rightarrow {\mathrm{e}}^{3{\mathrm{\beta }}^{-1}}=e(3+2\sqrt{2}) \end{gathered}$$