# If the solve the problem

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:

$y=a x^{n+1}+b x^{-n}$

$\frac{d y}{d x}=(n+1) a x^{n}+(-n) b x^{-n-1}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{n}(\mathrm{n}+1) \mathrm{ax}^{\mathrm{n}-1}+(-\mathrm{n})(-\mathrm{n}-1) \mathrm{bx}^{-\mathrm{n}-2}$

$x^{2} \frac{d^{2} y}{d x^{2}}=x^{2}\left\{n(n+1) a x^{n-1}+(-n)(-n-1) b x^{-n-2}\right\}=\lambda y$

$\lambda y=n(n+1) a x^{n-1+2}+n(n+1) b x^{-n-2+2}$

$\lambda y=n(n+1)\left[a x^{\wedge}(n+1)+b x^{\wedge}(-n)\right]$

$\lambda y=n(n+1)$

$\lambda=n(n+1)$