Question:

<p><strong>Step 1:</strong> Rewrite the equation in standard form by dividing by sin x:</p><p>$$\frac{dy}{dx} + \frac{\cos x}{\sin x}y = \frac{4x}{\sin x}$$</p><p><strong>Step 2:</strong> Recognize that the left side is the derivative of a product. Notice that:</p><p>$$\frac{d}{dx}[y \sin x] = y\cos x + \sin x\frac{dy}{dx}$$</p><p>This matches our original equation exactly!</p><p><strong>Step 3:</strong> Integrate both sides:</p><p>$$\int d[y\sin x] = \int 4x\,dx$$</p><p>$$y\sin x = 2x^2 + C$$</p><p><strong>Step 4:</strong> Apply the initial condition $y\left(\frac{\pi}{2}\right) = 0$:</p><p>$$0 \cdot \sin\frac{\pi}{2} = 2\left(\frac{\pi}{2}\right)^2 + C$$</p><p>$$0 = \frac{\pi^2}{2} + C \implies C = -\frac{\pi^2}{2}$$</p><p><strong>Step 5:</strong> Find $y\left(\frac{\pi}{6}\right)$:</p><p>$$y\sin\frac{\pi}{6} = 2\left(\frac{\pi}{6}\right)^2 - \frac{\pi^2}{2}$$</p><p>$$y \cdot \frac{1}{2} = \frac{\pi^2}{18} - \frac{\pi^2}{2} = \frac{\pi^2 - 9\pi^2}{18} = -\frac{8\pi^2}{18} = -\frac{4\pi^2}{9}$$</p><p>$$y\left(\frac{\pi}{6}\right) = -\frac{8\pi^2}{9}$$</p>