Let y : (-∞, ∞) → (0, ∞) be the solution of the differential equation $\frac{dy}{dx}=\frac{e^{5x}{y}^3+y^3}{e^x+e^x{y}^4}$ satisfying $y\left(0\right)=\frac{1}{\sqrt{2}}$. Then the value of y(log_e 2) is

The correct option is $\sqrt{\frac{7+\sqrt{53}}{2}}$.

Step 1: Simplify and separate variables in the differential equation.
The given first-order ordinary differential equation is:

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

Factor out common terms in the numerator and the denominator:

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

Rearrange terms to separate the variables $y$ and $x$:

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

Simplify each side by dividing term-by-term:

$(y^{-3}+y)dy=(e^{4x}+e^{-x})dx$

Step 2: Integrate both sides.
Integrating both sides with respect to their corresponding variables:

$\int (y^{-3}+y)dy=\int (e^{4x}+e^{-x})dx$

Evaluating the integrals yields:

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

Multiply the entire equation by 4 to clear the fractions:

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

where $K=4C$ is an arbitrary constant.

Step 3: Apply the initial condition.
We are given that $y\left(0\right)=\frac{1}{\sqrt{2}}$. Substitute $x=0$ and $y=\frac{1}{\sqrt{2}}$ into the equation:

Since $y^2=\frac{1}{2}$:

$2\left(\frac{1}{2}\right)-\frac{2}{1/2}=e^0-4e^0+K$

$1-4=1-4+K$

$-3=-3+\mathrm{K}\Rightarrow K=0$

Thus, the relationship simplifies to:

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

Step 4: Find the value of $y\left({\log }_{e}2\right)$.
Substitute $x={\log }_{e}2$ into the equation:

$e^{4x}=e^{4{\log }_{e}2}=e^{{\log }_e\left(2^4\right)}=16$

$e^{-x}=e^{-{\log }_{e}2}=e^{{\log }_e(1/2)}=\frac{1}{2}$

Substituting these exponential values into the equation:

$2y^2-\frac{2}{y^2}=16-4\left(\frac{1}{2}\right)=16-2=14$

Divide by 2:

$y^2-\frac{1}{y^2}=7$

Let $t=y^2$ (where $t>0$ since $y>0$):

$t-\frac{1}{t}=7\Rightarrow t^2-7t-1=0$

Solve this quadratic equation using the quadratic formula:

$t=\frac{-(-7)\pm \sqrt{(-7{)}^2-4(1\left)\right(-1)}}{2\left(1\right)}=\frac{7\pm \sqrt{49+4}}{2}=\frac{7\pm \sqrt{53}}{2}$

Since $t=y^2>0$ and $\sqrt{53}>7$, we discard the negative root to preserve positivity:

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

Taking the positive square root (since $y>0$):

$y\left({\log }_{e}2\right)=\sqrt{\frac{7+\sqrt{53}}{2}}$