Question:
If \(y = \left(\frac{2}{\pi}x - 1\right)\mathrm{cosec}\,x\) is the solution of the differential equation,
\(\frac{dy}{dx} + p(x)\,y = \frac{2}{\pi}\mathrm{cosec}\,x\), \(0 \lt x \lt \frac{\pi}{2}\),
then the function \(p(x)\) is equal to :
If \(y = \left(\frac{2}{\pi}x - 1\right)\mathrm{cosec}\,x\) is the solution of the differential equation,
\(\frac{dy}{dx} + p(x)\,y = \frac{2}{\pi}\mathrm{cosec}\,x\), \(0 \lt x \lt \frac{\pi}{2}\),
then the function \(p(x)\) is equal to :
\(\frac{dy}{dx} + p(x)\,y = \frac{2}{\pi}\mathrm{cosec}\,x\), \(0 \lt x \lt \frac{\pi}{2}\),
then the function \(p(x)\) is equal to :
Show Hint
Write \(y\sin x = \frac{2x}{\pi} - 1\) and differentiate it; compare with the given linear equation.
Updated On: Nov 5, 2026
- \(\cot x\)
- \(\sec x\)
- \(\tan x\)
- \(\mathrm{cosec}\,x\)
Show Solution
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The Correct Option is A
Solution and Explanation
Given \(y = \left(\frac{2x}{\pi} - 1\right)\mathrm{cosec}\,x\), so \(y\sin x = \frac{2x}{\pi} - 1\).
Differentiate both sides w.r.t. \(x\): \(\frac{dy}{dx}\sin x + y\cos x = \frac{2}{\pi}\).
Divide by \(\sin x\) (non-zero for \(0 \lt x \lt \frac{\pi}{2}\)): \(\frac{dy}{dx} + (\cot x)\,y = \frac{2}{\pi}\mathrm{cosec}\,x\).
Comparing with \(\frac{dy}{dx} + p(x)y = \frac{2}{\pi}\mathrm{cosec}\,x\) gives \(p(x) = \cot x\).
Answer: (1) \(\cot x\)
Differentiate both sides w.r.t. \(x\): \(\frac{dy}{dx}\sin x + y\cos x = \frac{2}{\pi}\).
Divide by \(\sin x\) (non-zero for \(0 \lt x \lt \frac{\pi}{2}\)): \(\frac{dy}{dx} + (\cot x)\,y = \frac{2}{\pi}\mathrm{cosec}\,x\).
Comparing with \(\frac{dy}{dx} + p(x)y = \frac{2}{\pi}\mathrm{cosec}\,x\) gives \(p(x) = \cot x\).
Answer: (1) \(\cot x\)
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