The general solution of the differential equation \(\displaystyle\frac{dy}{dx} - y\tan x = 2x\sec x\), \(-\displaystyle\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(y = (x^2 + C)\sec x\)
This is a linear equation with \(P = -\tan x\) and \(Q = 2x\sec x\). On the given interval \(\cos x \gt 0\), so the integrating factor is
\( e^{-\int \tan x\,dx} = e^{-\log \sec x} = \cos x \)
The solution is
\( \displaystyle y\cos x = \int 2x\sec x \cdot \cos x\,dx = \int 2x\,dx \)
\( y\cos x = x^2 + C \)
Dividing by \(\cos x\):
\( y = (x^2 + C)\sec x \)
Why the other options are incorrect
Option (1) \(y = (x^2 + C)\cos x\) — Incorrect. \(y\cos x = x^2 + C\) is multiplied by \(\cos x\) instead of divided by it.
Option (2) \(y = x^2\sec x + C\) — Incorrect. The constant must also be multiplied by \(\sec x\) when dividing by the integrating factor.
Option (4) \(y = (x^2 + C)\,\mathrm{cosec}\,x\) — Incorrect. This would come from the integrating factor \(\sin x\), which belongs to \(P = \cot x\), not to \(P = -\tan x\).