Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>Let \(y = y(x)\) be the solution of the differential equation \(\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x\), \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(y(0) = 1\). Then <strong>(JEE Main 2019)</strong></p>
<p>(a) \(y'\left(\frac{\pi}{4}\right) - y'\left(-\frac{\pi}{4}\right) = \pi - 2\)</p>
<p>(b) (remaining options not provided in text)</p>
<p>(c) (remaining options not provided in text)</p>
<p>(d) (remaining options not provided in text)</p>
Step-by-Step Solution
Key Concept: Use the integrating factor method with $\mu(x) = \sec x$, then differentiate the solution to find y' at specific points.
<p><strong>Step 1:</strong> Recognize this as a linear first-order differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \tan x$ and $Q(x) = 2x + x^2 \tan x$.</p><p><strong>Step 2:</strong> Find the integrating factor $\mu(x) = e^{\int \tan x \, dx} = e^{-\ln|\cos x|} = \sec x$.</p><p><strong>Step 3:</strong> Multiply the equation by $\sec x$ and solve by integration.</p><p><strong>Step 4:</strong> Use the initial condition $y(0) = 1$ to find the particular solution.</p><p><strong>Step 5:</strong> Compute $y'(\pi/4)$ and $y'(-\pi/4)$ using the original differential equation and verify the relationship.</p><p>∴ Answer is A.</p>
Correct Answer: A