Let y : (-∞, ∞) arrow (0, ∞) be the solution of the differential...

The given differential equation is

$\dfrac{dy}{dx} = \dfrac{e^{5x} y^3 + y^3}{e^x + e^x y^4}$

Factoring the numerator and denominator, we get

$\dfrac{dy}{dx} = \dfrac{y^3 (e^{5x} + 1)}{e^x (1 + y^4)}$

Separating the variables $x and y$, we obtain

$\dfrac{1 + y^4}{y^3} dy = \dfrac{e^{5x} + 1}{e^x} dx$

\left( \dfrac{1}{y^3} + y \right) dy = (e^{4x} + e^{-x}) dx$<br><br>Integrating both sides yields<br><br>$\int \left( y^{-3} + y \right) dy = \int (e^{4x} + e^{-x}) dx$<br><br>$-\dfrac{1}{2y^2} + \dfrac{y^2}{2} = \dfrac{e^{4x}}{4} - e^{-x} + C$<br><br>Multiplying by$2$, we get<br><br>$y^2 - \dfrac{1}{y^2} = \dfrac{e^{4x}}{2} - 2e^{-x} + 2C$<br><br>Using the initial condition$y(0) = \dfrac{1}{\sqrt{2}}$, we substitute x = 0 and$y = \dfrac{1}{\sqrt{2}}$:<br><br>\left(\dfrac{1}{\sqrt{2}}\right)^2 - \dfrac{1}{\left(\dfrac{1}{\sqrt{2}}\right)^2} = \dfrac{e^0}{2} - 2e^0 + 2C$

$\dfrac{1}{2} - 2 = \dfrac{1}{2} - 2 + 2C$

This gives $2C = 0$, so $C = 0$. The equation becomes

$y^2 - \dfrac{1}{y^2} = \dfrac{e^{4x}}{2} - 2e^{-x}$

To find $y(\log_e 2)$, we substitute $x = \log_e 2$ into the equation. We have $e^{4x} = e^{4 \log_e 2} = 16 and e^{-x} = e^{-\log_e 2} = \dfrac{1}{2}$.

$y^2 - \dfrac{1}{y^2} = \dfrac{16}{2} - 2\left(\dfrac{1}{2}\right)$

$y^2 - \dfrac{1}{y^2} = 8 - 1 = 7$

Let $y^2 = t$. Since $y \in (0, \infty)$, $t > 0$.

$t - \dfrac{1}{t} = 7$

$t^2 - 7t - 1 = 0$

Solving for $t$ using the quadratic formula gives

$t = \dfrac{7 \pm \sqrt{49 - 4(1)(-1)}}{2} = \dfrac{7 \pm \sqrt{53}}{2}$

Since $t = y^2 > 0$, we must choose the positive root:

$y^2 = \dfrac{7 + \sqrt{53}}{2}$

Taking the positive square root since $y > 0$, we get

$y = \sqrt{\dfrac{7 + \sqrt{53}}{2}}$