Question

Let \(y(x)\) be the solution of the differential equation \[ x\frac{dy}{dx}= y + x^2\cot x, \quad y\!\left(\frac{\pi}{2}\right)=\frac{\pi}{2}. \] The value of \(6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right)\) equals:

• A. \(-\pi\)
• B. \(-2\pi\)
• C. \(\pi\)
• D. \(2\pi\)

Hint:

For equations of the form \(x\dfrac{dy}{dx}-y=f(x)\), divide by \(x\) first and look for an integrating factor \(\frac{1}{x}\).

Answer 1

The Correct Option is A

To solve the given differential equation, we follow these steps:

1. The differential equation is given by: \(x\frac{dy}{dx} = y + x^2\cot x\).
2. This can be rewritten in the standard linear form as: \(\frac{dy}{dx} - \frac{y}{x} = x\cot x\).
3. The integrating factor (IF) for this linear differential equation is: \(e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x}\).
4. Multiplying the entire equation by the integrating factor, we get: \(\frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = \cot x\) which simplifies to: \(\frac{d}{dx}\left(\frac{y}{x}\right) = \cot x\).
5. Integrating both sides with respect to \(x\): \(\frac{y}{x} = \int \cot x \, dx = \ln|\sin x| + C\).
6. Thus, the general solution is: \(y = x(\ln|\sin x| + C)\).
7. Use the initial condition \(y\!\left(\frac{\pi}{2}\right) = \frac{\pi}{2}\) to find \(C\): \(\frac{\pi}{2} = \frac{\pi}{2}\ln|\sin\left(\frac{\pi}{2}\right)| + \frac{\pi}{2}C\).
8. The equation simplifies as \(\ln 1 = 0\), so: \(C = 1\).
9. Therefore, \(y = x(\ln|\sin x| + 1)\).
10. Now, calculate \(6y\!\left(\frac{\pi}{6}\right) - 8y\!\left(\frac{\pi}{4}\right)\):
    • For \(y\left(\frac{\pi}{6}\right)\): \(y\left(\frac{\pi}{6}\right) = \frac{\pi}{6} \left( \ln \left| \frac{1}{2} \right| + 1 \right) = \frac{\pi}{6} \left(\ln \frac{1}{2} + 1\right)\).
    • For \(y\!\left(\frac{\pi}{4}\right)\): \(y\!\left(\frac{\pi}{4}\right) = \frac{\pi}{4} \left( \ln \left| \frac{\sqrt{2}}{2} \right| + 1 \right) = \frac{\pi}{4} \left(\ln \frac{\sqrt{2}}{2} + 1\right)\).
11. Compute the required expression:
    • \(6y\!\left(\frac{\pi}{6}\right) = 6 \times \frac{\pi}{6} \left(\ln \frac{1}{2} + 1\right) = \pi(\ln \frac{1}{2} + 1)\)
    • \(8y\!\left(\frac{\pi}{4}\right) = 8 \times \frac{\pi}{4} \left(\ln \frac{\sqrt{2}}{2} + 1\right) = 2\pi\left(-\ln 2 + 1\right)\)
    • Combining, we have: \(6y\!\left(\frac{\pi}{6}\right) - 8y\!\left(\frac{\pi}{4}\right) = \pi(\ln \frac{1}{2} + 1) - 2\pi(-\ln 2 + 1)\) 
    • Simplify: \(\pi(-\ln 2 + 1) + 2\pi(\ln 2 - 1) = -\pi\)

Thus, the final answer is: \(-\pi\).