Let y(x) be the solution of the differential equation

2x² dy + (e^y $-$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :

$2{x^2}dy + ({e^y} - 2x)dx = 0$<br><br>$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$<br><br>$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} = - {1 \over {2{x^2}}} \Rightarrow $$ Put ${e^{ - y}} = z$<br><br>$${{ - dz} \over {dx}} - {z \over x} = - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}$$<br><br>$d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c$<br><br>$x{e^{ - y}} = {1 \over 2}{\log _e}x + c$, passes through (e, 1)<br><br>$\Rightarrow C = {1 \over 2}$<br><br>$x{e^{ - y}} = {{{{\log }_e}ex} \over 2}$<br><br>${e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2$