QJEEM19Ma9fc9: JEE Main 2019 Mathematics - Class 12 JEEM19Ma9fc9 Solution: If e y + xy = e, the ordered pair at x = 0 is equal to :

Question

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If e^y + xy = e, the ordered pair

\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right)

at x = 0 is equal to :

Accepted Answer

y = 1

\Rightarrow

x = 0

{e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0

\Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}

\Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0

x = 0, y = 1

\Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0

\Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}

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