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

Step 1: Rewrite the differential equation \[ x\frac{dy}{dx}-y=-x^2\cot x \] This is a linear differential equation of the form: \[ \frac{dy}{dx}-\frac{1}{x}y=-x\cot x \]
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 \[ \frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=-\cot x \] Left-hand side becomes: \[ \frac{d}{dx}\left(\frac{y}{x}\right) \] Thus: \[ \frac{d}{dx}\left(\frac{y}{x}\right)=-\cot x \]
Step 4: Integrate \[ \frac{y}{x}=-\ln(\sin x)+C \] \[ y=x\bigl(C-\ln(\sin x)\bigr) \]
Step 5: Use the given condition \[ y\!\left(\frac{\pi}{2}\right)=\frac{\pi^2}{2} \] Since \(\sin\frac{\pi}{2}=1\): \[ \frac{\pi^2}{2}=\frac{\pi}{2}C \Rightarrow C=\pi \] Thus: \[ y=x\left(\pi-\ln(\sin x)\right) \]
Step 6: Evaluate required expression \[ y\!\left(\frac{\pi}{6}\right)=\frac{\pi}{6}\left(\pi-\ln\frac12\right) \] \[ y\!\left(\frac{\pi}{4}\right)=\frac{\pi}{4}\left(\pi-\ln\frac{\sqrt2}{2}\right) \] Using: \[ \ln\frac12=-\ln2,\quad \ln\frac{\sqrt2}{2}=-\frac12\ln2 \] Compute: \[ 6y\!\left(\frac{\pi}{6}\right)-8y\!\left(\frac{\pi}{4}\right) =3\pi \] Final Answer: \[ \boxed{3\pi} \]