If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to

<p>Given,</p> <p>$(1 + {e^{2x}}){{dy} \over {dx}} + 2(1 + {y^2}){e^x} = 0$</p> <p>$\Rightarrow {{dy} \over {dx}} = {{ - 2(1 + {y^2}){e^x}} \over {1 + {e^{2x}}}}$</p> <p>$$ \Rightarrow \int {{{dy} \over {1 + {y^2}}} = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} } $$</p> <p>$\Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}}$</p> <p>Now,</p> <p>Let ${e^x} = t$</p> <p>$\Rightarrow {e^x}dx = dt$</p> <p>$\Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2dt} \over {1 + {t^2}}}}$</p> <p>$\Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}(t) + C$</p> <p>$\Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}({e^x}) + C$ [Putting value of t]</p> <p>Given, y(0) = 0 means when x = 0 the y = 0</p> <p>$\therefore$ ${\tan ^{ - 1}}(0) = - 2{\tan ^{ - 1}}({e^o}) + C$</p> <p>$\Rightarrow 0 = - 2 \times {\pi \over 4} + C$</p> <p>$\Rightarrow C = {\pi \over 2}$</p> <p>$\therefore$ ${\tan ^{ - 1}}(y) = 2{\tan ^{ - 1}}({e^x}) + {\pi \over 2}$</p> <p>$$ \Rightarrow y = \tan \left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right)$$</p> <p>Differentiating both sides, we get</p> <p>$$y' = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) - 2\,.\,{{{e^x}} \over {1 + {e^{2x}}}}$$</p> <p>$$ = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) \times {{ - 2{e^x}} \over {1 + {e^{2x}}}}$$</p> <p>$\therefore$ $$y'(0) = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^o}) + {\pi \over 2}} \right) \times {{ - 2{e^o}} \over {1 + {e^o}}}$$</p> <p>$$ = {\sec ^2}\left( { - 2 \times {\pi \over 4} + {\pi \over 2}} \right) \times {{ - 2} \over 2}$$</p> <p>$= {\sec ^2}(0)x - 1$</p> <p>$= 1 \times 1$</p> <p>$= - 1$</p> <p>And</p> <p>$$y\left( {\log _e^{\sqrt 3 }} \right) = \tan \left( { - 2{{\tan }^{ - 1}}\left( {{e^{\log _e^{\sqrt 3 }}}} \right) + {\pi \over 2}} \right)$$</p> <p>$= \tan \left( { - 2{{\tan }^{ - 1}}(\sqrt 3 ) + {\pi \over 2}} \right)$</p> <p>$= \tan \left( { - 2 \times {\pi \over 3} + {\pi \over 2}} \right)$</p> <p>$= \tan \left( { - {\pi \over 6}} \right)$</p> <p>$= - \tan \left( {{\pi \over 6}} \right)$</p> <p>$= - {1 \over {\sqrt 3 }}$</p> <p>$\therefore$ $6\left( {y'(0) + {{\left( {y(\log _e^{\sqrt 3 })} \right)}^2}} \right)$</p> <p>$= 6\left( { - 1 + {{\left( {{{ - 1} \over {\sqrt 3 }}} \right)}^2}} \right)$</p> <p>$= 6\left( { - 1 + {1 \over 3}} \right) = 6 \times {{ - 2} \over 3} = - 4$</p>