JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 7)
Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0$$
then, the minimum value of $$y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$$ is equal to :
then, the minimum value of $$y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$$ is equal to :
$$\left( {2 - \sqrt 3 } \right) - {\log _e}2$$
$$\left( {2 + \sqrt 3 } \right) + {\log _e}2$$
$$\left( {1 + \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$
$$\left( {1 - \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$
Explanation
$${{dy - dx} \over {{e^{y - x}}}} = xdx$$
$$ \Rightarrow {{dy - dx} \over {{e^{y - x}}}} = xdx$$
$$ \Rightarrow - {e^{x - y}} = {{{x^2}} \over 2} + c$$
At x = 0, y = 0 $$\Rightarrow$$ c = $$-$$1
$$ \Rightarrow {e^{x - y}} = {{2 - {x^2}} \over 2}$$
$$ \Rightarrow y = x - \ln \left( {{{2 - {x^2}} \over 2}} \right)$$
$$ \Rightarrow {{dy} \over {dx}} = 1 + {{2x} \over {2 - {x^2}}} = {{2 + 2x - {x^2}} \over {2 - {x^2}}}$$
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So minimum value occurs at $$x = 1 - \sqrt 3 $$
$$y\left( {1 - \sqrt 3 } \right) = \left( {1 - \sqrt 3 } \right) - \ln \left( {{{2 - \left( {4 - 2\sqrt 3 } \right)} \over 2}} \right)$$
$$ = \left( {1 - \sqrt 3 } \right) - \ln \left( {\sqrt 3 - 1} \right)$$
$$ \Rightarrow {{dy - dx} \over {{e^{y - x}}}} = xdx$$
$$ \Rightarrow - {e^{x - y}} = {{{x^2}} \over 2} + c$$
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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