Question 441 of 490MediumMCQ
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
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Official solution
Given,
\[(1 + {e^{2x}}){{dy} \over {dx}} + 2(1 + {y^2}){e^x} = 0\]
\[ \Rightarrow {{dy} \over {dx}} = {{ - 2(1 + {y^2}){e^x}} \over {1 + {e^{2x}}}}\]
\[ \Rightarrow \int {{{dy} \over {1 + {y^2}}} = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} } \]
\[ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2{e^x}dx} \over {(1 + {e^{2x}})}}} \]
Now,
Let ex=t
\[ \Rightarrow {e^x}dx = dt\]
\[ \Rightarrow {\tan ^{ - 1}}(y) = \int {{{ - 2dt} \over {1 + {t^2}}}} \]
\[ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}(t) + C\]
\[ \Rightarrow {\tan ^{ - 1}}(y) = - 2{\tan ^{ - 1}}({e^x}) + C\] [Putting value of t]
Given, y(0) = 0 means when x = 0 the y = 0
\[\therefore\] \[{\tan ^{ - 1}}(0) = - 2{\tan ^{ - 1}}({e^o}) + C\]
\[ \Rightarrow 0 = - 2 \times {\pi \over 4} + C\]
\[ \Rightarrow C = {\pi \over 2}\]
\[\therefore\] \[{\tan ^{ - 1}}(y) = 2{\tan ^{ - 1}}({e^x}) + {\pi \over 2}\]
\[ \Rightarrow y = \tan \left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right)\]
Differentiating both sides, we get
\[y' = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) - 2\,.\,{{{e^x}} \over {1 + {e^{2x}}}}\]
\[ = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^x}) + {\pi \over 2}} \right) \times {{ - 2{e^x}} \over {1 + {e^{2x}}}}\]
\[\therefore\] \[y'(0) = {\sec ^2}\left( { - 2{{\tan }^{ - 1}}({e^o}) + {\pi \over 2}} \right) \times {{ - 2{e^o}} \over {1 + {e^o}}}\]
\[ = {\sec ^2}\left( { - 2 \times {\pi \over 4} + {\pi \over 2}} \right) \times {{ - 2} \over 2}\]
\[ = {\sec ^2}(0)x - 1\]
\[ = 1 \times 1\]
=−1
And
\[y\left( {\log _e^{\sqrt 3 }} \right) = \tan \left( { - 2{{\tan }^{ - 1}}\left( {{e^{\log _e^{\sqrt 3 }}}} \right) + {\pi \over 2}} \right)\]
\[ = \tan \left( { - 2{{\tan }^{ - 1}}(\sqrt 3 ) + {\pi \over 2}} \right)\]
\[ = \tan \left( { - 2 \times {\pi \over 3} + {\pi \over 2}} \right)\]
\[ = \tan \left( { - {\pi \over 6}} \right)\]
\[ = - \tan \left( {{\pi \over 6}} \right)\]
\[ = - {1 \over {\sqrt 3 }}\]
\[\therefore\] \[6\left( {y'(0) + {{\left( {y(\log _e^{\sqrt 3 })} \right)}^2}} \right)\]
\[ = 6\left( { - 1 + {{\left( {{{ - 1} \over {\sqrt 3 }}} \right)}^2}} \right)\]
\[ = 6\left( { - 1 + {1 \over 3}} \right) = 6 \times {{ - 2} \over 3} = - 4\]
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