Let y = y(x), y > 0, be a solution curve of the differential...

Given,

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

$⇒ \dfrac{dy}{dx} = \dfrac{x}{1 + {x}^{2}}y - \dfrac{{y}^{2}}{1 + {x}^{2}}$

$⇒ \dfrac{- 1}{{y}^{2}}\dfrac{dy}{dx} + \dfrac{x}{1 + {x}^{2}} \cdot \dfrac{1}{y} = \dfrac{1}{1 + {x}^{2}}$

Now let $\dfrac{1}{y} = t$, we get,

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

So, the equation becomes,

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

Now finding, $IF = {e}^{∫ \dfrac{x}{1 + {x}^{2}}dx} = \sqrt{1 + {x}^{2}}$

So, solution of the differential equation is given by,

$t\sqrt{1 + {x}^{2}} = ∫ \dfrac{\sqrt{1 + {x}^{2}}}{1 + {x}^{2}}dx$

$⇒ \dfrac{1}{y}\sqrt{1 + {x}^{2}} = ∫ \dfrac{1}{\sqrt{1 + {x}^{2}}}dx$

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

$∵ y\left(0\right) = 1 ⇒ C = 1$

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

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

$⇒ \dfrac{3}{\beta} = \ln \left|2\sqrt{2} + 3\right| + 1$

$⇒ \beta = \dfrac{3}{1 + \ln \left|2\sqrt{2} + 3\right|}$

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

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

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