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Let the solution curve $$y = y(x)$$ of the differential equation $$\left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]x{{dy} \over {dx}} = x + \left[ {{x \over {\sqrt {{x^2} - {y^2}} }} + {e^{{y \over x}}}} \right]y$$ pass through the points (1, 0) and (2$$\alpha$$, $$\alpha$$), $$\alpha$$ > 0. Then $$\alpha$$ is equal to

JEE · Math · previous-year question

  1. A.$${1 \over 2}\exp \left( {{\pi \over 6} + \sqrt e - 1} \right)$$correct
  2. B.$${1 \over 2}\exp \left( {{\pi \over 6} + e - 1} \right)$$
  3. C.$$\exp \left( {{\pi \over 6} + \sqrt e + 1} \right)$$
  4. D.$$2\exp \left( {{\pi \over 3} + \sqrt e - 1} \right)$$

Answer

A. $${1 \over 2}\exp \left( {{\pi \over 6} + \sqrt e - 1} \right)$$

Explanation

$$\left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){{dy} \over {dx}} = 1 + \left( {{1 \over {\sqrt {1 - {{{y^2}} \over {{x^2}}}} }} + {e^{{y \over x}}}} \right){y \over x}$$ Putting y = tx $$\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)\left( {t + x{{dt} \over {dx}}} \right) = 1 + \left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right)t$$ $$ \Rightarrow x\left( {{1 \over {\sqrt {1 - {t^2}} }} + {e^t}} \right){{dt} \over {dx}} = 1$$ $$ \Rightarrow {\sin ^{ - 1}}t + {e^t} = \ln x + C$$ $$ \Rightarrow {\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + C$$ at x = 1, y = 0 So, $$0 + {e^0} = 0 + C \Rightarrow C = 1$$ at $$(2\alpha ,\alpha )$$ $${\sin ^{ - 1}}\left( {{y \over x}} \right) + {e^{y/x}} = \ln x + 1$$ $$ \Rightarrow {\pi \over 6} + {e^{{1 \over 2}}} - 1 = \ln (2\alpha )$$ $$ \Rightarrow \alpha = {1 \over 2}{e^{\left( {{\pi \over 6} + {e^{{1 \over 2}}} - 1} \right)}}$$

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