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Let y(x) be the solution of the differential equation 2x2 dy + (ey $$-$$ 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :

JEE · Math · previous-year question

  1. A.0
  2. B.2
  3. C.loge 2correct
  4. D.loge (2e)

Answer

C. loge 2

Explanation

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

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