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

Concept: The question involves differentiation of functions containing powers of \(x\). We repeatedly apply the power rule: \[ \frac{d}{dx}(x^m)=mx^{m-1} \] After finding the second derivative, we multiply by \(x^2\) and simplify to express the result in terms of the original function \(y\).

Step 1: Writing the given function clearly.
We are given: \[ y=ax^{n+1}+bx^{-n} \] where \(a\) and \(b\) are constants.

Step 2: Finding the first derivative.
Differentiate term by term: \[ \frac{dy}{dx} = a(n+1)x^n+b(-n)x^{-n-1} \] Simplifying: \[ \frac{dy}{dx} = a(n+1)x^n-bnx^{-n-1} \]

Step 3: Finding the second derivative.
Differentiate again: \[ \frac{d^2y}{dx^2} = a(n+1)n x^{n-1} -bn(-n-1)x^{-n-2} \] Simplify carefully: \[ \frac{d^2y}{dx^2} = n(n+1)ax^{n-1} +n(n+1)bx^{-n-2} \] Factor out the common factor: \[ \frac{d^2y}{dx^2} = n(n+1)\left(ax^{n-1}+bx^{-n-2}\right) \]

Step 4: Multiplying by \(x^2\).
Now multiply both sides by \(x^2\): \[ x^2\frac{d^2y}{dx^2} = x^2\cdot n(n+1)\left(ax^{n-1}+bx^{-n-2}\right) \] Distribute \(x^2\): \[ x^2\frac{d^2y}{dx^2} = n(n+1)\left(ax^{n+1}+bx^{-n}\right) \] Notice that: \[ ax^{n+1}+bx^{-n}=y \] Therefore, \[ x^2\frac{d^2y}{dx^2} = n(n+1)y \] Hence, \[ \boxed{x^2\frac{d^2y}{dx^2}=n(n+1)y} \]