Solution of Differential Equation and Value Evaluation

To find the solution of the given differential equation, we start by separating the variables.

The given differential equation is: $\frac{dy}{dx} = \frac{e^{5x}y^3 + y^3}{e^x + e^x y^4}$

Factoring out $y^3$ in the numerator and $e^x$ in the denominator: $\frac{dy}{dx} = \frac{y^3\left(e^{5x} + 1\right)}{e^x\left(1 + y^4\right)}$

Separating the variables $x$ and $y$: $\left(\frac{1 + y^4}{y^3}\right) dy = \left(\frac{e^{5x} + 1}{e^x}\right) dx$

Simplifying both sides gives: $\left(y^{-3} + y\right) dy = \left(e^{4x} + e^{-x}\right) dx$

Integrating both sides with respect to their corresponding variables: $\int \left(y^{-3} + y\right) dy = \int \left(e^{4x} + e^{-x}\right) dx$

$-\frac{1}{2y^2} + \frac{y^2}{2} = \frac{e^{4x}}{4} - e^{-x} + C$

Using the initial condition $y(0) = \frac{1}{\sqrt{2}}$, we substitute $x = 0$ and $y^2 = \frac{1}{2}$: $-\frac{1}{2\left(\frac{1}{2}\right)} + \frac{\frac{1}{2}}{2} = \frac{e^0}{4} - e^0 + C$

$-1 + \frac{1}{4} = \frac{1}{4} - 1 + C$

$-\frac{3}{4} = -\frac{3}{4} + C \implies C = 0$

Thus, the implicit solution to the differential equation is: $\frac{y^2}{2} - \frac{1}{2y^2} = \frac{e^{4x}}{4} - e^{-x}$

Multiplying the entire equation by $2$: $y^2 - \frac{1}{y^2} = \frac{e^{4x}}{2} - 2e^{-x}$

Now, we evaluate $y(x)$ at $x = \log_e 2$. For $x = \log_e 2$, we have: $e^x = 2, \quad e^{-x} = \frac{1}{2}, \quad \text{and} \quad e^{4x} = (e^x)^4 = 2^4 = 16$

Substituting these values into the solution: $y^2 - \frac{1}{y^2} = \frac{16}{2} - 2\left(\frac{1}{2}\right) = 8 - 1 = 7$

Let $u = y^2$ (where $u > 0$ as $y: (-\infty, \infty) \to (0, \infty)$): $u - \frac{1}{u} = 7 \implies u^2 - 7u - 1 = 0$

Using the quadratic formula to solve for $u$: $u = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(-1)}}{2} = \frac{7 \pm \sqrt{53}}{2}$

Since $u = y^2 > 0$, we take the positive root: $y^2 = \frac{7 + \sqrt{53}}{2}$

Taking the positive square root for $y$: $y(\log_e 2) = \sqrt{\frac{7 + \sqrt{53}}{2}}$

Hence, the correct option is B.