Let $y:(-\infty, \infty) \rightarrow(0, \infty)$ be the solution of the differential equation $\frac{dy}{dx}=\frac{{e}^{5x}{y}^{3}+{y}^{3}}{{e}^{x}+{e}^{x}{y}^{4}}$ satisfying $y\left(0\right)=\frac{1}{\sqrt{2}}$. Then the value of $y\left(\log _e 2\right)$ is [JEE Advanced 2026]

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YESPYQ · Accepted Answer

Correct answer: b) $\sqrt{\frac{7+\sqrt{53}}{2}}$. $\dfrac{dy}{dx}=\dfrac{y^3(e^{5x}+1)}{e^x(1+y^4)}$ $\dfrac{1+y^4}{y^3}\,dy=(e^{4x}+e^{-x})\,dx$ $\int\left(y+\dfrac1{y^3}\right)dy=\int(e^{4x}+e^{-x})dx$ $\dfrac{y^2}{2}-\dfrac1{2y^2}=\dfrac{e^{4x}}4-e^{-x}+C$ $y(0)=\dfrac1{\sqrt2}$ $\dfrac{1/2}{2}-\dfrac1{2(1/2)}=\dfrac14-1+C$ $\dfrac14-1=\dfrac14-1+C$ $C=0$ $\dfrac{y^2}{2}-\dfrac1{2y^2}=\dfrac{e^{4x}}4-e^{-x}$ At $x=\ln2$, $\dfrac{y^2}{2}-\dfrac1{2y^2}=4-\dfrac12=\dfrac72$ $y^2-\dfrac1{y^2}=7$ $y^4-7y^2-1=0$ $y^2=\dfrac{7+\sqrt{53}}2$ $y=\sqrt{\dfrac{7+\sqrt{53}}2}$

Let \(y:(-\infty, \infty) \rightarrow(0, \infty)\) be the solution of the differential equation \(\frac{dy}{dx}=\frac{{e…

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