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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 :

JEE · Math · previous-year question

  1. A.$${1 \over {1 - {{\log }_e}2}}$$
  2. B.$${1 \over {1 + {{\log }_e}2}}$$correct
  3. C.$${{ - 1} \over {1 + {{\log }_e}2}}$$
  4. D.$${1 + {{\log }_e}2}$$

Answer

B. $${1 \over {1 + {{\log }_e}2}}$$

Explanation

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

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