If y = y(x) is the solution of the differential equation (1 + e^2x)di…

The correct answer is Option (3) → \(\displaystyle\frac{1}{2}\)

Separating the variables:

    \( \displaystyle \frac{dy}{1 + y^2} = \frac{e^{x}}{1 + e^{2x}}\,dx \)

For the right side, put \(t = e^{x}\), so \(dt = e^{x}\,dx\):

    \( \displaystyle \int \frac{e^{x}}{1 + e^{2x}}\,dx = \int \frac{dt}{1 + t^2} = \tan^{-1} (e^{x}) \)

So

    \( \tan^{-1} y = \tan^{-1} (e^{x}) + C \)

Using \(y(0) = 0\): \(0 = \tan^{-1} 1 + C\), so

    \( \displaystyle C = -\frac{\pi}{4} \)

At \(x = \log_e 3\), \(e^{x} = 3\):

    \( \displaystyle y = \tan\left(\tan^{-1} 3 - \frac{\pi}{4}\right) \)

Using \(\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}\):

    \( \displaystyle y = \frac{3 - 1}{1 + 3 \cdot 1} = \frac{1}{2} \)

The constant must be \(-\frac{\pi}{4}\), not \(+\frac{\pi}{4}\): with \(+\frac{\pi}{4}\) the value would be \(\frac{3 + 1}{1 - 3} = -2\), and \(y(0)\) would not be 0.