Differential Equations
Linear Differential Equations and Integrating Factor
Grade 12
Question:
<p>If \(y = y(x)\) is the solution of the differential equation \(\frac{dy}{dx} = (\tan x - y) \sec^2 x\), \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(y(0) = 0\), then \(y\left(-\frac{\pi}{4}\right)\) is equal to <strong>(JEE Main 2019)</strong></p>
<p>(a) \(1\)</p>
<p>(b) \(-e\)</p>
<p>(c) \(2 + \frac{1}{e}\)</p>
<p>(d) \(e^{-2}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a linear first-order ODE and use the integrating factor method with $\mu(x) = e^{\tan x}$.
<p><strong>Step 1:</strong> Rewrite the differential equation as $\frac{dy}{dx} + y\sec^2 x = \tan x \sec^2 x$, which is a linear first-order ODE.</p><p><strong>Step 2:</strong> Identify the integrating factor as $\mu(x) = e^{\int \sec^2 x \, dx} = e^{\tan x}$.</p><p><strong>Step 3:</strong> Multiply both sides by the integrating factor and integrate.</p><p><strong>Step 4:</strong> Apply the initial condition $y(0) = 0$ to find the constant of integration.</p><p><strong>Step 5:</strong> Evaluate at $x = -\frac{\pi}{4}$ to get $y\left(-\frac{\pi}{4}\right) = -e$.</p><p>∴ Answer is B.</p>
Correct Answer: B