If a curve$y = f\left(x\right)$ passing through the point $(1, 2)$ is the solution of the differential equation 

$2x^{2}dy=\left(2xy+y^2\right)dx$ then $f\left(\frac{1}{2}\right)$ is  equal to 

The given differential equation is 

$\begin{array}{l}2x^{2}dy=\left(2xy+y^2\right)dx\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{2x^2}\\ \Rightarrow -\frac{1}{y^2}\frac{dy}{dx}+\left(\frac{1}{x}\right)\frac{1}{y}=\frac{-1}{2x^2}\\ \Rightarrow \frac{dv}{dx}+\frac{1}{x}v=-\frac{1}{2x^2},\text{ where }v=\frac{1}{y}\end{array}$

This is a linear differential equation with $\text{ I. F. }{e}^{\int \frac{1}{x}dx}=e^{{\log }_{e}x}=x$

Multiplying both sides by $x$and integrating, we obtain 

$vx=\frac{1}{2}\log x+C\Rightarrow \frac{x}{y}=-\frac{1}{2}\log x+C$            (i)

This is the required curve. It passes through $(1, 2)$. 

$\therefore \frac{1}{2}=C$

Putting $C=\frac{1}{2}$ in (i), we obtain 

$\begin{array}{l}\frac{x}{y}=-\frac{1}{2}\log x+\frac{1}{2}\Rightarrow \frac{2x}{y}=1-\log x\Rightarrow y=\frac{2x}{1-\log x}\\ \therefore f\left(x\right)=\frac{2x}{1-\log x}\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{1+{\log }_{c}2}\end{array}$