If (dy)/(dx) = xy x ^ 2 + y ^ 2 ; y(1) = 1; then a value of x...

Put $y = vx$

$\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$

$v + x\dfrac{dv}{dx} = \dfrac{v{x}^{2}}{{x}^{2} + {v}^{2}{x}^{2}}$

$⇒ \dfrac{1 + {v}^{2}}{{v}^{3}}dv = - \dfrac{1}{x}dx$

$⇒ ∫ \left(\dfrac{1}{{v}^{3}} + \dfrac{1}{v}\right)dv = ∫ - \dfrac{1}{x}dx$

$⇒ \dfrac{- 1}{2}\dfrac{1}{{v}^{2}} + \ln v = - \ln x + c$

$⇒ - \dfrac{{x}^{2}}{2{y}^{2}} = - \ln y + c$

When $x = 1, y = 1$ then

$- \dfrac{1}{2} = c$

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

$⇒ {x}^{2} = {e}^{2}\left(3\right)$