If a curve $y=f(x)$, passing through the point $(1,2)$, is the solution of the differential equation,
$2 x^{2} d y=\left(2 x y+y^{2}\right) d x$, then $f\left(\frac{1}{2}\right)$ is equal to :
Correct Option: , 2
$2 x^{2} d y=\left(2 x y+y^{2}\right) d x$
$\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$ \{Homogeneous D.E.\}
$\left\{\begin{array}{l}\operatorname{let} y=x t \\ \Rightarrow \frac{d y}{d x}=t+x \frac{d t}{d x}\end{array}\right\}$
$\Rightarrow t+x \frac{d t}{d x}=\frac{2 x^{2} t+x^{2} t^{2}}{2 x^{2}}$
$\Rightarrow \mathrm{t}+\mathrm{x} \frac{\mathrm{dt}}{\mathrm{dx}}=\mathrm{t}+\frac{\mathrm{t}^{2}}{2}$
$\Rightarrow 2 \int \frac{\mathrm{dt}}{\mathrm{t}^{2}}=\int \frac{\mathrm{dx}}{\mathrm{x}}$
$\Rightarrow 2\left(-\frac{1}{t}\right)=\ell \mathrm{n}(\mathrm{x})+\mathrm{C}\left\{\right.$ Put $\left.\mathrm{t}=\frac{\mathrm{y}}{\mathrm{x}}\right\}$
$\Rightarrow-\frac{2 x}{y}=\ell n x+C\left\{\begin{array}{l}\text { Put } x=1 \& y=2 \\ \text { then we get } C=-1\end{array}\right\}$
$\Rightarrow \frac{-2 x}{y}=\ell n(x)-1$
$\Rightarrow y=\frac{2 x}{1-\ell n x}$
$\Rightarrow f(x)=\frac{2 x}{1-\log _{e} x}$
so, $f\left(\frac{1}{2}\right)=\frac{1}{1+\log _{e} 2}$
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