If \(y=\mathrm{ax}^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}\), then \(x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}=\)

\(y=a x^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{a}(\mathrm{n}+1) x^{\mathrm{n}+1-1}+\mathrm{b}(-\mathrm{n}) x^{-\mathrm{n}-1} \\
& \frac{\mathrm{~d} y}{\mathrm{~d} x}=\mathrm{a}(\mathrm{n}+1) x^{\mathrm{n}}-\mathrm{bn} x^{-\mathrm{n}-1}
\end{aligned}\)
Again differentiating w.r.t. \(x\), we get
\(\begin{aligned}
\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} & =a(n+1)^n x^{\mathrm{n}-1}-b n(-n-1) x^{-n-1-1} \\
& =a(n+1) n x^{n-1}+b n(n+1) x^{-n-2} \\
& =n(n+1) \frac{a x^n}{x}+n(n+1) b \frac{x^{-n}}{x^2}
\end{aligned}\)
\(\begin{array}{ll} & \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{n}(\mathrm{n}+1)}{x^2}\left[\mathrm{a} x^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}\right] \\ \therefore \quad & x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}=\mathrm{n}(\mathrm{n}+1) y\end{array}\)