Question

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\\( \frac { d y } { d \theta } = 4 y ^ { 2 } \sec ^ { 2 } ( 2 \theta ) \quad y \left( \frac { \pi } { 8 } \
ight) = 1 \\)

Explanation:

Step1: Separate variables

Separate the variables in the differential equation \(\frac{dy}{d\theta}=4y^{2}\sec^{2}(2\theta)\). We get \(\frac{dy}{y^{2}} = 4\sec^{2}(2\theta)d\theta\).

Step2: Integrate both sides

Integrate \(\int y^{- 2}dy=\int4\sec^{2}(2\theta)d\theta\).
For the left - hand side, using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\int y^{-2}dy=-\frac{1}{y}+C_1\).
For the right - hand side, let \(u = 2\theta\), then \(du=2d\theta\) and \(\int4\sec^{2}(2\theta)d\theta=4\times\frac{1}{2}\int\sec^{2}(u)du\). Since \(\int\sec^{2}(x)dx=\tan(x)+C\), we get \(2\tan(2\theta)+C_2\).
So, \(-\frac{1}{y}=2\tan(2\theta)+C\).

Step3: Use the initial condition \(y(\frac{\pi}{8}) = 1\)

Substitute \(\theta=\frac{\pi}{8}\) and \(y = 1\) into \(-\frac{1}{y}=2\tan(2\theta)+C\).
\(-1=2\tan(\frac{\pi}{4})+C\). Since \(\tan(\frac{\pi}{4}) = 1\), we have \(-1=2\times1+C\), then \(C=-3\).
The particular solution is \(-\frac{1}{y}=2\tan(2\theta)-3\), or \(y=\frac{1}{3 - 2\tan(2\theta)}\).

Step4: Evaluate \(y(\frac{3\pi}{8})\)

Substitute \(\theta=\frac{3\pi}{8}\) into \(y=\frac{1}{3 - 2\tan(2\theta)}\).
\(2\theta=\frac{3\pi}{4}\), and \(\tan(\frac{3\pi}{4})=-1\).
\(y(\frac{3\pi}{8})=\frac{1}{3-2\times(-1)}=\frac{1}{5}\).

Answer:

\(\frac{1}{5}\)