Let the solution curve of the differential equation xdy = ( x ^ 2 +...

Given,

$xdy = \left(\sqrt{{x}^{2} + {y}^{2}} + y\right)dx$

$⇒ xdy - ydx = \sqrt{{x}^{2} + {y}^{2}}dx$

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

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

Now integrating both side we get,

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

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

$⇒ \dfrac{y + \sqrt{{y}^{2} + {x}^{2}}}{x} = cx$

$⇒ y + \sqrt{{y}^{2} + {x}^{2}} = c{x}^{2}$

Now given when $x = 1, y = 0 ⇒ 0 + 1 = c ⇒ c = 1$

So equation of curve is $y + \sqrt{{x}^{2} + {y}^{2}} = {x}^{2}$

Now at $x = 2, y = \alpha$

So putting the value in curve we get, $2 + \sqrt{4 + {\alpha}^{2}} = 4$

$⇒ 4 + {\alpha}^{2} = 16 + {\alpha}^{2} = 8\alpha$

$⇒ \alpha = \dfrac{3}{2}$