#### Please solve RD Sharma class 12 chapter 11 Higher Order Derivatives exercise very short answer type question 1 maths textbook solution

$\lambda =n(n+1)$

Hint:

Double differentiate $y$  with respect to $x$ to get;

$\frac{d^{2}y}{dx^{2}}\: then \; substitute\; the\; val\! ue\; in\; x\frac{d^{2}y}{dx^{2}}=\lambda y \; to \; get\; \lambda .$

Given:

$y=ax^{n+1}+bx^{-n}$

Explanation:

It is given that

$y=ax^{n+1}+bx^{-n}\; \; \; \; \; \; \; ......(1)$

\begin{aligned} &\frac{d y}{d x}=a(n+1) x^{n+1-1}+b(-n) x^{-n-1} \quad\left[\frac{d}{d p} p^{9}=a p^{a-1}\right]\\ &\frac{d y}{d x}=a(n+1) x^{n}-b n x^{-(n+1)}\\ &\text { Again diff. above w.r.to } x\\ &\frac{d^{2} y}{d x^{2}}=n \cdot a(n+1) x^{n-1}-b n[-(n+1)] x^{-(n+1)-1}\\ &\frac{d^{2} y}{d x^{2}}=a n(n+1) x^{n-1}-b n(n+1) x^{-(n+2)} \ldots . .(2) \end{aligned}

We have,

$x^{2}\frac{d^{2}y}{dx^{2}}=\lambda y \; \; \; \; \; \; .....(3)$

Using (1), (2) & (3) becomes

\begin{aligned} &\begin{aligned} &x^{2}\left[\operatorname{an}(n+1) x^{n-1}+b n(n+1) x^{-(n+2)}\right] \\ &=\lambda\left[a x^{n+1}+b x^{-n}\right] \\ &n(n+1)\left[a x^{n+1}+b x^{-n}\right]=\lambda\left[a x^{n+1}+b x^{-n}\right] \\ &\lambda=n(n+1) \end{aligned}\\ &\text { Hence, the value of } \lambda \text { is } n(n+1) \text { . } \end{aligned}