Question:
Solution of the equation
\[
\cos^2 x \frac{dy}{dx} - (\tan 2x)\,y = \cos^4 x,\quad |x|<\frac{\pi}{4},
\]
where \(y\!\left(\frac{\pi}{6}\right)=\frac{3\sqrt{3}}{8}\), is given by:
Solution of the equation
\[
\cos^2 x \frac{dy}{dx} - (\tan 2x)\,y = \cos^4 x,\quad |x|<\frac{\pi}{4},
\]
where \(y\!\left(\frac{\pi}{6}\right)=\frac{3\sqrt{3}}{8}\), is given by:
Show Hint
Convert trigonometric expressions into \(\tan x\) form to simplify linear differential equations.
Updated On: Apr 16, 2026
- \(y\,\frac{\tan 2x}{1-\tan^2 x} = 0\)
- \(y(1-\tan^2 x) = C\)
- \(y = \sin 2x + C\)
- \(y = \frac{1}{2}\cdot\frac{\sin 2x}{1-\tan^2 x}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept: First-order linear differential equation.
Step 1: \[ \frac{dy}{dx} - \frac{\tan 2x}{\cos^2 x}\, y = \cos^2 x \]
Step 2: \[ \tan 2x = \frac{2\tan x}{1 - \tan^2 x}, \quad \cos^2 x = \frac{1}{1 + \tan^2 x} \]
Step 3: \[ y = \frac{1}{2} \cdot \frac{\sin 2x}{1 - \tan^2 x} \]
Step 4:
\( x = \frac{\pi}{6} \) satisfies the given condition.
Step 1: \[ \frac{dy}{dx} - \frac{\tan 2x}{\cos^2 x}\, y = \cos^2 x \]
Step 2: \[ \tan 2x = \frac{2\tan x}{1 - \tan^2 x}, \quad \cos^2 x = \frac{1}{1 + \tan^2 x} \]
Step 3: \[ y = \frac{1}{2} \cdot \frac{\sin 2x}{1 - \tan^2 x} \]
Step 4:
\( x = \frac{\pi}{6} \) satisfies the given condition.
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