If \(y = y(x)\), \(0 \lt x \lt \pi\), is the solution of the differential equation \(\displaystyle\frac{dy}{dx} + y\cot x = 2\cos x\) with \(y\left(\displaystyle\frac{\pi}{2}\right) = 2\), then \(y\left(\displaystyle\frac{\pi}{6}\right)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{5}{2}\)
This is a linear equation with \(P = \cot x\) and \(Q = 2\cos x\). On \(0 \lt x \lt \pi\), \(\sin x \gt 0\), so the integrating factor is
\( e^{\int \cot x\,dx} = e^{\log \sin x} = \sin x \)
So
\( \displaystyle y\sin x = \int 2\sin x\cos x\,dx \)
\( y\sin x = \sin^2 x + C \)
Using \(y\left(\frac{\pi}{2}\right) = 2\):
\( 2 \times 1 = 1 + C \Rightarrow C = 1 \)
\( y = \sin x + \mathrm{cosec}\,x \)
At \(x = \frac{\pi}{6}\):
\( \displaystyle y = \frac{1}{2} + 2 = \frac{5}{2} \)
Why the other options are incorrect
Option (2) \(\frac{3}{2}\) — Incorrect. This is \(2 - \frac{1}{2}\), which takes \(y = \mathrm{cosec}\,x - \sin x\); that function gives \(y\left(\frac{\pi}{2}\right) = 0\), not 2.
Option (3) \(\frac{9}{2}\) — Incorrect. This takes \(C = 2\), the given value of y, but \(2 = 1 + C\) gives \(C = 1\).
Option (4) \(1\) — Incorrect. This takes \(y = 2\sin x\), which leaves out the term \(\mathrm{cosec}\,x\) and does not satisfy the equation.