Question:

<div class='solution'><p><strong>Step 1:</strong> Rewrite: $\dfrac{d}{dx}(y\sin x) = \sin x\,y' + y\cos x$.</p> <p>So the equation is: $\dfrac{d}{dx}(y\sin x) = -(2x + x^2\cos x)$.</p> <p><strong>Step 2:</strong> Integrate both sides:</p> <p>$$y\sin x = -\int(2x + x^2\cos x)\,dx$$</p> <p>$\int 2x\,dx = x^2$. For $\int x^2\cos x\,dx$: integrate by parts twice:</p> <p>$= x^2\sin x - 2\int x\sin x\,dx = x^2\sin x - 2(-x\cos x + \sin x) = x^2\sin x + 2x\cos x - 2\sin x$.</p> <p>$$y\sin x = -(x^2 + x^2\sin x + 2x\cos x - 2\sin x) + C$$</p> <p><strong>Step 3:</strong> At $x=0$: $0 = 0 + C \implies C = 0$.</p> <p>At $x = \pi/2$, $\sin(\pi/2)=1$: $y = -(\pi^2/4 + \pi^2/4 + 0 - 2) + 2 = -\pi^2/4$.</p> <p>Wait — simplify: $y(\pi/2) = -\pi^2/4$. <strong>Answer: (A)</strong> $-\pi^2/4$.</p> <p class='key-concept'>🔑 Key Concept: When you see $\sin x\cdot y' + y\cos x$, immediately recognise it as $\frac{d}{dx}(y\sin x)$. This converts the equation to direct integration.</p> <p class='trap-warning'>⚠️ Trap: Dividing through by $\sin x$ and treating as a standard linear ODE — valid but messy. The exact derivative shortcut is far cleaner.</p></div>