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

$$ \begin{aligned} & \left(1+x^2\right) d y=y(x-y) d x \\\\ & y(0)=1 \cdot y(2 \sqrt{2})=\beta \\\\ & \frac{d y}{d x}=\frac{y x-y^2}{1+x^2} \\\\ & \frac{d y}{d x}+y\left(\frac{-x}{1+x^2}\right)=\left(\frac{-1}{1+x^2}\right) y^2 \\\\ & \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{y}\left(\frac{-x}{1+x^2}\right)=\frac{-1}{1+x^2} \\\\ & \text { put } \frac{1}{y}=t \text { then } \frac{-1}{y^2} \frac{d y}{d x}=\frac{d t}{d x} \end{aligned} $$ <br/><br/>$$ \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \frac{\mathrm{x}}{1+\mathrm{x}^2}=\frac{1}{1+\mathrm{x}^2} $$ <br/><br/>$$ \text { I.F }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\frac{1}{2} \ln \left(1+\mathrm{x}^2\right)}=\sqrt{1+\mathrm{x}^2} $$ <br/><br/>$t \sqrt{1+x^2}=\int \frac{\sqrt{1+x^2}}{1+x^2} d x$ <br/><br/>$$ \begin{aligned} & \frac{1}{y} \sqrt{1+x^2}=\int \frac{1}{\sqrt{1+x^2}} d x \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+C \\\\ & \because y(0)=1 \Rightarrow C=1 \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+1 \\\\ & \text { For } y=2 \sqrt{2} \\\\ & \frac{3}{y}=\ln |2 \sqrt{2}+3|+1 \end{aligned} $$ <br/><br/>$$ \begin{gathered} y=\beta=\frac{3}{1+\ln |2 \sqrt{2}+3|} \\\\ \Rightarrow 3 \beta^{-1}=1+\ln |2 \sqrt{2}+3| \end{gathered} $$ <br/><br/>Isolating the term $e^{3 \beta^{-1}}$, we get : <br/><br/>$e^{3 \beta^{-1}} = e^{1+\ln |2 \sqrt{2}+3|}.$ <br/><br/>This can be simplified using the rule $e^{a+b} = e^a e^b$ to : <br/><br/>$e^{3 \beta^{-1}} = e \cdot e^{\ln |2 \sqrt{2}+3|}.$ <br/><br/>Since $e^{\ln(x)} = x$ for any $x$, this simplifies to : <br/><br/>$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3|.$ <br/><br/>Using the given value of $\beta$, which is $\frac{3}{1+\ln |2 \sqrt{2}+3|}$, we find : <br/><br/>$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3| = e (2\sqrt{2} + 3),$ <br/><br/>Since $2\sqrt{2}+3$ is positive and so the absolute value does not affect the result.