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

Step 1: Put the equation in linear form
\[ x\frac{dy}{dx}=y+x^2\cot x \;\Rightarrow\; \frac{dy}{dx}-\frac{1}{x}y=x\cot x \] This is a linear first-order differential equation.
Step 2: Find the integrating factor
\[ \text{I.F.}=e^{\int -\frac{1}{x}\,dx}=\frac{1}{x} \]
Step 3: Multiply throughout by the integrating factor
\[ \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=\cot x \] \[ \Rightarrow \frac{d}{dx}\!\left(\frac{y}{x}\right)=\cot x \]
Step 4: Integrate
\[ \frac{y}{x}=\int \cot x\,dx=\ln(\sin x)+C \] \[ y=x\bigl(\ln(\sin x)+C\bigr) \]
Step 5: Use the given condition
\[ y\!\left(\frac{\pi}{2}\right)=\frac{\pi}{2} \] \[ \frac{\pi}{2}=\frac{\pi}{2}\bigl(\ln 1 + C\bigr) \Rightarrow C=1 \] \[ \boxed{y=x\bigl(\ln(\sin x)+1\bigr)} \]
Step 6: Evaluate required expression
\[ y\!\left(\frac{\pi}{6}\right) =\frac{\pi}{6}\!\left(\ln\frac12+1\right) =\frac{\pi}{6}(1-\ln2) \] \[ y\!\left(\frac{\pi}{4}\right) =\frac{\pi}{4}\!\left(\ln\frac{\sqrt2}{2}+1\right) =\frac{\pi}{4}\!\left(1-\frac12\ln2\right) \] \[ 6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right) = \pi(1-\ln2)-2\pi\!\left(1-\frac12\ln2\right) \] \[ = \pi-\pi\ln2-2\pi+\pi\ln2 = -\pi \] \[ \boxed{-\pi} \]