Question

For \(n\in\mathbb{N}\), if \[ y=ax^{n+1}+bx^{-n}, \] then \[ x^2\frac{d^2y}{dx^2} \] is equal to:

• A. \((n-1)y\)
• B. \(n(n-1)y\)
• C. \(n(n+1)y\)
• D. \((n+1)y\)

Hint:

For powers of \(x\), remember: \[ x^2\frac{d^2}{dx^2}(x^m)=m(m-1)x^m \] This shortcut is very useful in Euler-Cauchy type expressions.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves successive differentiation (finding the second derivative).
The function given is a linear combination of two power terms of \( x \).
We apply the power rule of differentiation repeatedly.
The final expression involves multiplying the second derivative by \( x^2 \) and simplifying it to relate it back to the original function \( y \).
Step 2: Key Formula or Approach:
Power Rule: \( \frac{d}{dx} x^k = kx^{k-1} \).
We need to find \( \frac{dy}{dx} \) then \( \frac{d^2y}{dx^2} \).
Step 3: Detailed Explanation:
The given function is:
\[ y = ax^{n+1} + bx^{-n} \]
Differentiating once with respect to \( x \):
\[ \frac{dy}{dx} = a(n + 1)x^{(n + 1) - 1} + b(-n)x^{-n - 1} \]
\[ \frac{dy}{dx} = a(n + 1)x^n - bnx^{-n - 1} \]
Differentiating again to find the second derivative:
\[ \frac{d^2y}{dx^2} = a(n + 1) \cdot n \cdot x^{n - 1} - bn \cdot (-n - 1) \cdot x^{-n - 2} \]
Simplifying the coefficients:
\[ \frac{d^2y}{dx^2} = n(n + 1)ax^{n - 1} + n(n + 1)bx^{-n - 2} \]
Factoring out the common term \( n(n + 1) \):
\[ \frac{d^2y}{dx^2} = n(n + 1) [ax^{n - 1} + bx^{-n - 2}] \]
Now, calculate \( x^2 \frac{d^2y}{dx^2} \):
\[ x^2 \frac{d^2y}{dx^2} = x^2 \cdot n(n + 1) [ax^{n - 1} + bx^{-n - 2}] \]
Distributing \( x^2 \) into the terms inside the bracket:
\[ x^2 \frac{d^2y}{dx^2} = n(n + 1) [ax^{n - 1 + 2} + bx^{-n - 2 + 2}] \]
\[ x^2 \frac{d^2y}{dx^2} = n(n + 1) [ax^{n + 1} + bx^{-n}] \]
Observe that the expression in the bracket is exactly \( y \).
\[ x^2 \frac{d^2y}{dx^2} = n(n + 1)y \]
Step 4: Final Answer:
The expression is \( n(n + 1)y \).
This matches Option (C).