Question:
If $y=a x^{n+1}+b x^{-n}$ and $x^{2} \frac{d^{2} y}{d x^{2}}=\lambda y$, then write the value of $\lambda$
Solution:
Given:
$y=a x^{n+1}+b x^{-n}$
$\frac{d y}{d x}=(n+1) a x^{n}+(-n) b x^{-n-1}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{n}(\mathrm{n}+1) \mathrm{ax}^{\mathrm{n}-1}+(-\mathrm{n})(-\mathrm{n}-1) \mathrm{bx}^{-\mathrm{n}-2}$
$x^{2} \frac{d^{2} y}{d x^{2}}=x^{2}\left\{n(n+1) a x^{n-1}+(-n)(-n-1) b x^{-n-2}\right\}=\lambda y$
$\lambda y=n(n+1) a x^{n-1+2}+n(n+1) b x^{-n-2+2}$
$\lambda y=n(n+1)\left[a x^{\wedge}(n+1)+b x^{\wedge}(-n)\right]$
$\lambda y=n(n+1)$
$\lambda=n(n+1)$