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

Let y:(-∞,∞)→(0,∞) be the solution of the differential equation dy/dx=(e⁵ˣy³+y³)/(eˣ+eˣy⁴), satisfying y(0)=1/√2. Then the value of y(logₑ2) is $\dfrac{dy}{dx}=\dfrac{e^{5x}y^{3}+y^{3}}{e^{x}+e^{x}y^{4}}=\dfrac{y^{3}\,(e^{5x}+1)}{e^{x}\,(1+y^{4})}$ $y(0)=\dfrac{1}{\sqrt{2}}$ $\text{Find } y(\ln 2)$ $(1)\ \sqrt{\tfrac{5+\sqrt{35}}{2}}, \quad (2)\ \sqrt{\tfrac{7+\sqrt{53}}{2}}$ $(3)\ \tfrac{7+\sqrt{53}}{2}, \quad (4)\ \tfrac{5+\sqrt{35}}{2}$ $1.\ \text{separate variables}$ $2.\ \text{integrate both sides}$ $3.\ \text{use } y(0); \ \text{then } x=\ln 2$ $\dfrac{1+y^{4}}{y^{3}}\,dy=\dfrac{e^{5x}+1}{e^{x}}\,dx$ $\left(y^{-3}+y\right)dy=\left(e^{4x}+e^{-x}\right)dx$ $\dfrac{y^{2}}{2}-\dfrac{1}{2y^{2}}=\dfrac{e^{4x}}{4}-e^{-x}+C$ $x=0,\ y^{2}=\tfrac{1}{2}:\ \text{LHS}=\tfrac{1}{4}-1=-\tfrac{3}{4}$ $\text{RHS}=\tfrac{1}{4}-1=-\tfrac{3}{4} \Rightarrow C=0$ $y^{2}-\dfrac{1}{y^{2}}=\dfrac{e^{4x}}{2}-2e^{-x}$ $x=\ln 2:\ e^{4x}=16,\quad e^{-x}=\tfrac{1}{2}$ $y^{2}-\dfrac{1}{y^{2}}=8-1=7$ $u-u^{-1}=7 \Rightarrow u^{2}-7u-1=0$ $u=\dfrac{7\pm\sqrt{53}}{2}$ $u=y^{2}>0 \Rightarrow y^{2}=\dfrac{7+\sqrt{53}}{2}$ $y=\sqrt{\dfrac{7+\sqrt{53}}{2}}$ $\boxed{y=\sqrt{\dfrac{7+\sqrt{53}}{2}} \ \Rightarrow\ \text{option }(2)}$