QJEEM21Mae143: JEE Main 2021 Mathematics - Class 12 JEEM21Mae143 Solution: Let y = y(x) be the solution of the differential equation then, …

Question

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Let y = y(x) be the solution of the differential equation

{{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 &lt; x &lt; \sqrt 2 ,y(0) = 0

then, the minimum value of

y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)

is equal to :

Accepted Answer

{{dy - dx} \over {{e^{y - x}}}} = xdx

\Rightarrow {{dy - dx} \over {{e^{y - x}}}} = xdx

\Rightarrow - {e^{x - y}} = {{{x^2}} \over 2} + c

At x = 0, y = 0

\Rightarrow

c =

-

1

\Rightarrow {e^{x - y}} = {{2 - {x^2}} \over 2}

\Rightarrow y = x - \ln \left( {{{2 - {x^2}} \over 2}} \right)

\Rightarrow {{dy} \over {dx}} = 1 + {{2x} \over {2 - {x^2}}} = {{2 + 2x - {x^2}} \over {2 - {x^2}}}

So minimum value occurs at

x = 1 - \sqrt 3

y\left( {1 - \sqrt 3 } \right) = \left( {1 - \sqrt 3 } \right) - \ln \left( {{{2 - \left( {4 - 2\sqrt 3 } \right)} \over 2}} \right)

= \left( {1 - \sqrt 3 } \right) - \ln \left( {\sqrt 3 - 1} \right)

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