Question

Let \(cos^{-1}\frac{y}{b}\)=\(log_e (\frac{x}{n})^n\), then Ay₂+By₁+Cy=0 is possible :  (where y₂=\(\frac{d^2y}{dx^2}\), y₁=\(\frac{dy}{dx}\))

• A. A=2, B=x², C=n
• B. A=x², B=x, C=n²
• C. A=x, B=2x, C=3n+1
• D. A=x², B=3x, C=2n

Answer 1

The Correct Option is B

Given: \(\cos^{-1}\left(\frac{y}{b}\right) = \log\left(\frac{x}{n}\right)^n = n \log\left(\frac{x}{n}\right) = n(\log x - \log n)\)

Taking cosine on both sides:

\(\frac{y}{b} = \cos(n(\log x - \log n))\) 

\(y = b \cos(n(\log x - \log n))\)

Differentiating with respect to x:

\(y_1 = \frac{dy}{dx} = b \cdot (-\sin(n(\log x - \log n))) \cdot n \cdot \frac{1}{x} = -\frac{bn}{x} \sin(n(\log x - \log n))\)

\(x y_1 = -bn \sin(n(\log x - \log n))\) (1)

Differentiating again with respect to x:

\(y_2 = \frac{d^2y}{dx^2}\)

Differentiating \(x y_1 = -bn \sin(n(\log x - \log n))\)

\(x y_2 + y_1 = -bn \cdot \cos(n(\log x - \log n)) \cdot n \cdot \frac{1}{x}\)

\(x y_2 + y_1 = -\frac{bn^2}{x} \cos(n(\log x - \log n))\)

\(x^2 y_2 + x y_1 = -bn^2 \cos(n(\log x - \log n))\)

From the original equation \(y = b \cos(n(\log x - \log n))\) therefore \(bn\cos = y\):

\(x^2 y_2 + x y_1 = -n^2 y\)

\(x^2 y_2 + x y_1 + n^2 y = 0\)

Comparing this with \(Ay_2 + By_1 + Cy = 0\), we get \(A = x^2\), \(B = x\), and \(C = n^2\).

Conclusion:

The correct option is \(A = x^2, B = x, C = n^2\)