Discussion
Let \(\phi: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \mathbb{R}\) be the solution of the differential equation
\[ (\cos x)\frac{dy}{dx} - y = 2y^{2}(\sin x - 1)\cos x \]
satisfying \(\phi(0) = \frac{1}{2}\). Then \(\phi\left(\frac{\pi}{4}\right) =\) ______.
Answer: D
\(\phi\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\).
Dividing by \(\cos x\) gives \(y' - y\sec x = 2y^{2}(\sin x - 1)\), a Bernoulli equation. Divide by \(y^{2}\) and put \(v = \frac{1}{y}\), so \(v' = -\frac{y'}{y^{2}}\):
\[ v' + v\sec x = 2(1 - \sin x). \]
The integrating factor is \(\sec x + \tan x = \frac{1 + \sin x}{\cos x}\), and it turns the right side into \(\frac{2(1 - \sin x)(1 + \sin x)}{\cos x} = 2\cos x\). Hence
\[ \begin{aligned} v(\sec x + \tan x) &= 2\sin x + C \\ v(0) = 2 &\Rightarrow C = 2 \\ v &= \frac{2(1 + \sin x)\cos x}{1 + \sin x} = 2\cos x \end{aligned} \]
So \(\phi(x) = \frac{1}{2}\sec x\), and \(\phi\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\). Substituting back confirms it: both sides equal \(\frac{\sin x - 1}{2\cos x}\).
\(\frac{1}{2}\) is the starting value \(\phi(0)\), and 2 is \(v(0) = \frac{1}{\phi(0)}\), the substituted variable rather than \(y\). \(\frac{3 + 2\sqrt{2}}{2} \approx 2.91\) is far too large for a solution that only grows like \(\frac{1}{2}\sec x\).
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