Let y = y x ,   y > 0 , be a solution curve of the differential equation 1 + x 2 d y = y x - y d x …

Given,

$\left(1+x^2\right)dy=y\left(x-y\right)dx$

$\Rightarrow \frac{dy}{dx}=\frac{x}{1+x^2}y-\frac{y^2}{1+x^2}$

$\Rightarrow \frac{-1}{y^2}\frac{dy}{dx}+\frac{x}{1+x^2}·\frac{1}{y}=\frac{1}{1+x^2}$

Now let $\frac{1}{y}=t$, we get,

$\frac{-1}{y^2}{y}^{\text{'}}=\frac{dt}{dx}$

So, the equation becomes,

$\frac{dt}{dx}+\frac{x}{1+x^2}t=\frac{1}{1+x^2}$

Now finding, $\mathrm{IF}=e^{\int \frac{x}{1+x^2}dx}=\sqrt{1+x^2}$

So, solution of the differential equation is given by,

$t\sqrt{1+x^2}=\int \frac{\sqrt{1+x^2}}{1+x^2}dx$

$\Rightarrow \frac{1}{y}\sqrt{1+x^2}=\int \frac{1}{\sqrt{1+x^2}}dx$

$\Rightarrow \frac{1}{y}\sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+C$

$\because y\left(0\right)=1\Rightarrow C=1$

$\Rightarrow \frac{1}{y}\sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+1$

Now for $x=2\sqrt{2}, y=\beta$ we get,

$\Rightarrow \frac{3}{\beta }=\ln \left|2\sqrt{2}+3\right|+1$

$\Rightarrow \beta =\frac{3}{1+\ln \left|2\sqrt{2}+3\right|}$

$\Rightarrow 3{\beta }^{-1}=1+\ln \left|2\sqrt{2}+3\right|$

$\Rightarrow e^{3\beta -1}=e·e^{\ln \left(3+2\sqrt{2}\right)}$

$\Rightarrow e^{3\beta -1}=e\left(3+2\sqrt{2}\right)$