If a curve y = f (x), passing through the point (1, 2), is the...

\dfrac{dy}{dx} = \dfrac{y}{x} + \dfrac{{y}^{2}}{2{x}^{2}}$</p><p>$⇒ {y}^{- 2}\dfrac{dy}{dx} - \dfrac{1}{y} \cdot \dfrac{1}{x} = \dfrac{1}{2{x}^{2}}$</p><p>Put$- \dfrac{1}{y} = t ⇒ \dfrac{1}{{y}^{2}}\dfrac{dy}{dx} = \dfrac{dt}{dx}$</p><p>$⇒ \dfrac{dt}{dx} + \left(\dfrac{1}{x}\right)t = \dfrac{1}{2{x}^{2}}$</p><p>This is a linear differential equation,</p><p>with Integrating Factor :${e}^{∫ \dfrac{1}{x}dx} = {e}^{\ln \left(x\right)} = x$</p><p>So, solution of the linear differential equation is$tx = ∫ \dfrac{1}{2{x}^{2}} \cdot xdx + C$</p><p>$⇒ - \dfrac{x}{y} = \dfrac{1}{2}\ln \left(x\right) + C$</p><p> The curve passes through \left(1, 2\right)$

$⇒ - \dfrac{1}{2} = \dfrac{1}{2}\ln \left(1\right) + C ⇒ C = - \dfrac{1}{2}$

Hence, the particular solution to the differential equation is $- \dfrac{x}{y} = \dfrac{1}{2}\ln \left(x\right) - \dfrac{1}{2}$

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

$⇒ f\left(\dfrac{1}{2}\right) = \dfrac{2 \times \dfrac{1}{2}}{1 - \ln \left(\dfrac{1}{2}\right)} = \dfrac{1}{1 + \ln \left(2\right)} = \dfrac{1}{1 + \log_{e}\left(2\right)}$