If a curve y=f(x), passing through the point


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 :

  1. (1) $\frac{1}{1+\log _{e} 2}$

  2. (2) $\frac{1}{1-\log _{e} 2}$

  3. (3) $1+\log _{e} 2$

  4. (4) $\frac{-1}{1+\log _{e} 2}$

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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