Question

If y = 2xⁿ⁺¹ + \(\frac{3}{x^n}\) then x²\(\frac{d^2y}{dx^2}\) is

• A. 6n(n + 1)y
• B. n(n + 1)y
• C. \(x\frac{dy}{dx}+y\)
• D. y

Answer 1

The Correct Option is B

If \(y = 2x^{n+1} + \frac{3}{x^n}\), then we need to find \(x^2 \frac{d^2y}{dx^2}\).

Rewrite y as: \(y = 2x^{n+1} + 3x^{-n}\)

First, find the first derivative \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1}\)

Next, find the second derivative \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 2(n+1)(n)x^{n-1} - 3n(-n-1)x^{-n-2}\)

\(\frac{d^2y}{dx^2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}\)

Now, multiply by \(x^2\):

\(x^2 \frac{d^2y}{dx^2} = x^2 (2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2})\)

\(x^2 \frac{d^2y}{dx^2} = 2n(n+1)x^{n+1} + 3n(n+1)x^{-n}\)

\(x^2 \frac{d^2y}{dx^2} = n(n+1)(2x^{n+1} + 3x^{-n})\)

\(x^2 \frac{d^2y}{dx^2} = n(n+1)(2x^{n+1} + \frac{3}{x^n})\)

Since \(y = 2x^{n+1} + \frac{3}{x^n}\), we have:

\(x^2 \frac{d^2y}{dx^2} = n(n+1)y\)

Therefore, the correct option is (B) \(n(n+1)y\).

Answer 2

Given $y = 2x^{n+1} + \frac{3}{x^n} = 2x^{n+1} + 3x^{-n}$. 

First derivative: $\frac{dy}{dx} = 2(n+1)x^n + 3(-n)x^{-n-1} = 2(n+1)x^n - 3nx^{-n-1}$. 

Second derivative: $$\frac{d^2y}{dx^2} = 2(n+1)(n)x^{n-1} - 3n(-n-1)x^{-n-2} = 2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}$$ $$x^2 \frac{d^2y}{dx^2} = x^2 (2n(n+1)x^{n-1} + 3n(n+1)x^{-n-2}) = 2n(n+1)x^{n+1} + 3n(n+1)x^{-n}$$ $$x^2 \frac{d^2y}{dx^2} = n(n+1)(2x^{n+1} + 3x^{-n}) = n(n+1)y$$