If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x²dy= (2xy + y²)dx, then $f\left( {{1 \over 2}} \right)$ is equal to :

$2{x^2}dy = \left( {2xy + {y^2}} \right)dx$<br><br> $\Rightarrow 2{x^2}{{dy} \over {dx}} = 2xy + {y^2}$<br><br> $$ \Rightarrow {{2{x^2}} \over {2{x^2}{y^2}}}{{dy} \over {dx}} = {{2xy} \over {2{x^2}{y^2}}} + {{{y^2}} \over {2{x^2}{y^2}}}$$<br><br> $$ \Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over x}{1 \over y} = {1 \over {2{x^2}}}$$<br><br> Let $- {1 \over y} = t$<br><br> $\Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}$<br><br> $\Rightarrow {{dt} \over {dx}} + t{1 \over x} = {1 \over {2{x^2}}}$<br><br> This is linear differentiatial equation.<br><br> $\therefore I.F = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}}$<br><br> $\therefore t.{e^{\ln x}} = \int {{1 \over {2{x^2}}}.{e^{\ln x}}dx}$<br><br> $\Rightarrow - {1 \over y}.x = \int {{1 \over {2{x^2}}}.xdx}$<br><br> $\Rightarrow - {x \over y} = {1 \over 2}\int {{{dx} \over x}}$<br><br> $\Rightarrow - {x \over y} = {1 \over 2}\ln x + c$<br><br> This curve passes through the point (1, 2)<br><br> $\therefore - {1 \over 2} = 0 + c$<br><br> $\Rightarrow c = - {1 \over 2}$<br><br> $\therefore - {x \over y} = {1 \over 2}\ln x - {1 \over 2}$<br><br> $\Rightarrow - {{2x} \over y} = \ln x - 1$<br><br> $\Rightarrow y = {{2x} \over {1 - \ln x}}$<br><br> $\Rightarrow f\left( x \right) = {{2x} \over {1 - \ln x}}$<br><br> So, $$f\left( {{1 \over 2}} \right) = {{2 \times {1 \over 2}} \over {1 - \ln \left( {{1 \over 2}} \right)}} = {1 \over {1 + {{\log }_e}2}}$$