Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>If \(y(x)\) satisfies the differential equation \(\dfrac{dy}{dx} = \sin 2x + 3y \cot x\) and \(y\left(\dfrac{\pi}{2}\right) = 2\), then which of the following statement(s) is (are) correct?</p>
<p>(a) \(y\left(\dfrac{\pi}{6}\right) = 0\)</p>
<p>(b) \(y'\left(\dfrac{\pi}{3}\right) = \dfrac{9 - 3\sqrt{2}}{2}\)</p>
<p>(c) \(y(x)\) increases in interval \(\left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)\)</p>
<p>(d) The value of definite integral \(\displaystyle\int_{-\pi/2}^{\pi/2} y(x)\,dx\) equal \(\pi\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a linear first-order ODE in standard form dy/dx - 3y·cot(x) = sin(2x). Use integrating factor e^(∫-3cot(x)dx) = e^(-3ln|sin(x)|) = 1/sin³(x) to convert the left side into a perfect derivative, then integrate and apply the initial condition.
<p><strong>Step 1:</strong> Rewrite the equation in standard linear form: dy/dx - 3y·cot(x) = sin(2x)</p><p><strong>Step 2:</strong> Find the integrating factor: μ(x) = e^(∫-3cot(x)dx) = e^(-3ln|sin(x)|) = csc³(x) = 1/sin³(x)</p><p><strong>Step 3:</strong> Multiply both sides by μ(x): d/dx[y·csc³(x)] = sin(2x)·csc³(x) = 2cos(x)·csc²(x)</p><p><strong>Step 4:</strong> Integrate both sides: y·csc³(x) = ∫2cos(x)·csc²(x)dx</p><p>Let u = sin(x), then: y·csc³(x) = -2csc(x) + C</p><p><strong>Step 5:</strong> Apply initial condition y(π/2) = 2: 2·(1)³ = -2(1) + C → C = 4</p><p><strong>Step 6:</strong> General solution: y·csc³(x) = -2csc(x) + 4, or y = sin³(x)[-2csc(x) + 4] = -2sin²(x) + 4sin³(x)</p><p><strong>Step 7:</strong> Verify y(π/2) = -2(1) + 4(1) = 2 ✓. Check monotonicity, extrema, and other properties as per given options.</p><p>∴ Answer: ABD</p>
Correct Answer: ABD