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: 1
$\frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}$
It is homogeneous differential equation.
$\therefore \quad$ Put $y=v x$
$\Rightarrow v+x \frac{d v}{d x}=v+\frac{v^{2}}{2} \Rightarrow \int 2 \frac{d v}{v^{2}}=\int \frac{d x}{x}$
$\Rightarrow \frac{-2}{v}=\log _{e} x+c \Rightarrow \frac{-2 x}{y}=\log _{e} x+c$
Put $x=1, y=2$, we get $c=-1$
$\Rightarrow \frac{-2 x}{y}=\log _{e} x-1$
Hence, put $x=\frac{1}{2} \Rightarrow y=\frac{1}{1+\log _{e} 2}$
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