Consider the differential equation \(\displaystyle\frac{dy}{dx} + y\tan x = \cos^2 x\), \(-\displaystyle\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\).
(A) Its integrating factor is \(\sec x\).
(B) It is a differential equation of order 1 and degree 2.
(C) Its general solution is \(y = \sin x\cos x + C\cos x\), where C is an arbitrary constant.
(D) For the solution with \(y(0) = 1\), the value of y at \(x = \displaystyle\frac{\pi}{3}\) is \(\displaystyle\frac{\sqrt{3} + 1}{4}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
This is a linear equation with \(P = \tan x\) and \(Q = \cos^2 x\). On \(-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\), \(\sec x \gt 0\), so the integrating factor is
\( e^{\int \tan x\,dx} = e^{\log \sec x} = \sec x \)
So
\( \displaystyle y\sec x = \int \cos^2 x\sec x\,dx = \int \cos x\,dx \)
\( y\sec x = \sin x + C \)
\( y = \sin x\cos x + C\cos x \)
With \(y(0) = 1\): \(1 = 0 + C\), so \(C = 1\). At \(x = \frac{\pi}{3}\):
\( \displaystyle y = \frac{\sqrt{3}}{2}\cdot\frac{1}{2} + \frac{1}{2} = \frac{\sqrt{3} + 2}{4} \)
The highest derivative is \(\frac{dy}{dx}\), with power 1.
So the integrating factor is \(\sec x\), the order and degree are both 1, the general solution is \(y = \sin x\cos x + C\cos x\), and the particular solution gives \(\frac{\sqrt{3} + 2}{4}\) at \(\frac{\pi}{3}\).
Checking the statements
(A) Its integrating factor is \(\sec x\). — Correct. \(e^{\int \tan x\,dx} = \sec x\).
(B) It is a differential equation of order 1 and degree 2. — Incorrect. \(\frac{dy}{dx}\) appears with power 1, so the degree is 1.
(C) Its general solution is \(y = \sin x\cos x + C\cos x\). — Correct. \(y\sec x = \sin x + C\).
(D) For the solution with \(y(0) = 1\), the value of y at \(x = \frac{\pi}{3}\) is \(\frac{\sqrt{3} + 1}{4}\). — Incorrect. \(C = 1\) gives \(\frac{\sqrt{3}}{4} + \frac{1}{2} = \frac{\sqrt{3} + 2}{4}\).