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


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

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

  3. $\frac{-1}{1+\log _{e} 2}$

  4. $1+\log _{e} 2$

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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