Discussion
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
Answer: B
The value is \(\sqrt{\frac{7 + \sqrt{53}}{2}}\).
Factor both sides and separate the variables:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{y^{3}(e^{5x} + 1)}{e^{x}(1 + y^{4})} \\ \left(\frac{1}{y^{3}} + y\right)dy &= \left(e^{4x} + e^{-x}\right)dx \end{aligned} \]
Integrating both sides:
\[ \begin{aligned} -\frac{1}{2y^{2}} + \frac{y^{2}}{2} \\ = \frac{e^{4x}}{4} - e^{-x} + C \end{aligned} \]
At \(x = 0\), \(y^{2} = \frac{1}{2}\): the left side is \(-1 + \frac{1}{4} = -\frac{3}{4}\) and the right side is \(\frac{1}{4} - 1 + C\), so \(C = 0\). At \(x = \log_{e} 2\), \(e^{4x} = 16\) and \(e^{-x} = \frac{1}{2}\). Put \(u = y^{2}\):
\[ \begin{aligned} \frac{u}{2} - \frac{1}{2u} &= 4 - \frac{1}{2} = \frac{7}{2} \\ u^{2} - 7u - 1 &= 0 \\ u &= \frac{7 + \sqrt{53}}{2} \end{aligned} \]
The positive root is taken because \(u = y^{2} \gt 0\), and \(y \gt 0\) by the stated range, so \(y = \sqrt{u}\).
\(\frac{7 + \sqrt{53}}{2}\) is \(y^{2}\), not \(y\): it forgets the square root. The pair built on \(\sqrt{35}\) does not satisfy \(u^{2} - 7u - 1 = 0\).
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