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If $${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$; y(1) = 1; then a value of x satisfying y(x) = e is :

JEE · Math · previous-year question

  1. A.$$\sqrt 2 e$$
  2. B.$${1 \over 2}\sqrt 3 e$$
  3. C.$${e \over {\sqrt 2 }}$$
  4. D.$$\sqrt 3 e$$correct

Answer

D. $$\sqrt 3 e$$

Explanation

$${{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}$$ Let y = vx $$ \therefore $$ $${{dy} \over {dx}}$$ = v + x.$${{dv} \over {dx}}$$ $$ \Rightarrow $$ v + x.$${{dv} \over {dx}}$$ = $${{x\left( {vx} \right)} \over {{x^2} + {v^2}{x^2}}}$$ = $${v \over {1 + {v^2}}}$$ $$ \Rightarrow $$ x.$${{dv} \over {dx}}$$ = $${v \over {1 + {v^2}}}$$ - v = $${{v - v - {v^3}} \over {1 + {v^2}}}$$ = $$ - {{{v^3}} \over {1 + {v^2}}}$$ $$ \Rightarrow $$ $$\int {{{1 + {v^2}} \over {{v^3}}}} dv = - \int {{{dx} \over x}} $$ $$ \Rightarrow $$ $$ - {1 \over {2{v^2}}} + \log v$$ = $$ - \log x + C$$ $$ \Rightarrow $$ $$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x + C$$ ......(1) putting x = 1, y = 1 we get $$ \Rightarrow $$ C = $$ - {1 \over 2}$$ From eq. (1) $$ - {1 \over 2}{{{x^2}} \over {{y^2}}} + \log \left( {{y \over x}} \right)$$ = $$ - \log x - {1 \over 2}$$ Put y = e $$ - {1 \over 2}{{{x^2}} \over {{e^2}}} + \log \left( {{e \over x}} \right)$$ = $$ - \log x - {1 \over 2}$$ $$ \Rightarrow $$ x2 = 3e2 $$ \Rightarrow $$ x = $$ \pm $$3$$\sqrt e $$

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