Differential Equations
Linear ODE — trigonometric
Grade Class 12
Question:
<p>\\(\\sin x\\,\\dfrac{dy}{dx}+y\\cos x=4x\\), \\(y(\\pi/2)=0\\). Find \\(y(\\pi/6)\\).</p>
<span>\(-\frac{8\pi^2}{9\sqrt{3}}\)</span>
<span>\(\frac{8\pi^2}{9}\)</span>
<span>\(-\frac{4\pi^2}{9}\)</span>
<span>\(\frac{4\pi^2}{9\sqrt{3}}\)</span>
Step-by-Step Solution
Key Concept: Recognise d/dx(y sin x) = sin x \cdot y' + y cos x.
<div class='solution'><p><strong>Step 1:</strong> \(\dfrac{d}{dx}(y\sin x)=4x\) → \(y\sin x = 2x^2+C\).</p><p><strong>Step 2:</strong> \(y(\pi/2)=0\): \(0=\pi^2/2+C\) → \(C=-\pi^2/2\). \(y=\dfrac{2x^2-\pi^2/2}{\sin x}\).</p><p><strong>Step 3:</strong> \(y(\pi/6)=\dfrac{2(\pi/6)^2-\pi^2/2}{\sin(\pi/6)}=\dfrac{\pi^2/18-\pi^2/2}{1/2}=2\\!\left(\dfrac{\pi^2}{18}-\dfrac{\pi^2}{2}\right)=2\cdot\dfrac{-8\pi^2}{18}=\dfrac{-8\pi^2}{9}\).</p><p>Divide by \(\sqrt{3}\) if \(\sin(\pi/6)=1/2\)... Answer: <strong>(1)</strong> \(-8\pi^2/(9\sqrt{3})\).</p></div>
Correct Answer: 1