Question-33 For n\in\mathbb{N} if y=ax^{n+1}+bx^{-n}, then x^2\frac{d^2y}{dx^2}=
(A) n(n-1)y
(B) (n-1)y
(C) n(n+1)y
(D) (n+1)y
Answer: (C)
Explanation:
y=ax^{n+1}+bx^{-n}Differentiating w.r.t. x, we get
\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}
Again differentiating w.r.t. x, we get
\frac{d^2y}{dx^2}=a(n+1)n x^{n-1}-bn(-n-1)x^{-n-1-1} =a(n+1)n x^{n-1}+bn(n+1)x^{-n-2} =n(n+1)\frac{ax^n}{x}+n(n+1)\frac{bx^{-n}}{x^2} \frac{d^2y}{dx^2}=\frac{n(n+1)}{x^2}[ax^{n+1}+bx^{-n}]∴ x^2\frac{d^2y}{dx^2}=n(n+1)y