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Question

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

2
$-$2
$-$4
$-$1

Solution

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