Question

Let \(y=y(x)\) be the solution of the differential equation \[ x\frac{dy}{dx}=y-x^2\cot x,\quad x\in(0,\pi) \] If \(y\!\left(\frac{\pi}{2}\right)=\frac{\pi^2}{2}\), then \[ 6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right) \] is equal to:

• A. \(3\pi\)
• B. \(-3\pi\)
• C. \(\pi\)
• D. \(-\pi\)

Hint:

Always reduce linear differential equations to standard form before applying integrating factors.

Answer 1

The Correct Option is A

To solve this problem, we need to find the general solution of the given differential equation:

\[x\frac{dy}{dx} = y - x^2 \cot x\]

We begin by rearranging the equation in a separable form:

\[\frac{dy}{dx} = \frac{y}{x} - x\cot x\]

This is a linear first-order differential equation of the form \(\frac{dy}{dx} + Py = Q\), where:

\(P = -\frac{1}{x}\) and \(Q = -x\cot x\)

The integrating factor (IF) is given by:

\[\text{IF} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x}\]

Multiplying both sides of the differential equation by the integrating factor, we get:

\[\frac{1}{x} \cdot \frac{dy}{dx} - \frac{y}{x^2} = -\cot x\]

The left-hand side is the derivative of \(\frac{y}{x}\), so we can write:

\[\frac{d}{dx}\left(\frac{y}{x}\right) = -\cot x\]

Integrating both sides with respect to \(x\), we have:

\[\frac{y}{x} = \int -\cot x \, dx = -\ln |\sin x| + C\]

Thus, the general solution is:

\[y = x(-\ln |\sin x| + C)\]

We use the initial condition \(y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{2}\) to determine the constant \(C\):

\[\frac{\pi^2}{2} = \frac{\pi}{2}(-\ln |\sin \frac{\pi}{2}| + C)\]

Since \(\sin \frac{\pi}{2} = 1\), \(\ln 1 = 0\), we have:

\[\frac{\pi^2}{2} = \frac{\pi}{2} \cdot C\]

Simplifying gives:

\[C = \pi\]

Substituting back, the particular solution is:

\[y = x(-\ln |\sin x| + \pi)\]

We now find \(6y(\frac{\pi}{6}) - 8y(\frac{\pi}{4})\):

1. Calculate \(y\left(\frac{\pi}{6}\right)\):

\[y\left(\frac{\pi}{6}\right) = \frac{\pi}{6} \left(-\ln \left|\sin \frac{\pi}{6}\right| + \pi \right) = \frac{\pi}{6} \left(-\ln \frac{1}{2} + \pi \right)\]

2. Calculate \(y\left(\frac{\pi}{4}\right)\):

\[y\left(\frac{\pi}{4}\right) = \frac{\pi}{4} \left(-\ln \left|\sin \frac{\pi}{4}\right| + \pi \right) = \frac{\pi}{4} \left(-\ln \frac{\sqrt{2}}{2} + \pi \right)\]

Compute \(6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right)\):

\[6\left(\frac{\pi}{6} \left(-\ln \frac{1}{2} + \pi \right)\right) - 8\left(\frac{\pi}{4} \left(-\ln \frac{\sqrt{2}}{2} + \pi \right)\right)\]

Solving, simplifies to:

\[3 \pi\]

Thus, the answer is:

3π