Question

If y = a x^{n + 1} + bx⁻ⁿ and $x^2\frac{d^{2}y}{d{x}^2}=\mathrm{\lambda }y$, then write the value of λ.

Answer 1

Here,
$$\begin{gathered} y=a{x}^{n+1}+b{x}^{-n} \\ \mathrm{and}{x}^2\frac{d^{2}y}{d{x}^2}=\lambda y \\ \mathrm{Now}, \\ \frac{\mathit{d}y}{\mathit{d}x}=a\left(n+1\right)x^n-bn{x}^{-n-1} \\ \mathrm{and}\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=an\left(n+1\right)x^{n-1}-bn\left(-n-1\right)x^{-n-2} \\ \mathrm{Now},x^2\frac{{\mathit{d}}^{\mathit{2}}y}{\mathit{d}{x}^{\mathit{2}}}=\lambda y\left[\mathrm{Given}\right] \\ \Rightarrow x^2\left[an\left(n+1\right)x^{n-1}+bn\left(n+1\right)x^{-n-2}\right]=\lambda \left(a{x}^{n+1}+b{x}^{-n}\right) \\ \Rightarrow an\left(n+1\right)x^{n+1}+bn\left(n+1\right)x^{-n}=\lambda \left(a{x}^{n+1}+b{x}^{-n}\right) \\ \Rightarrow n\left(n+1\right)\left(a{x}^{n+1}+b{x}^{-n}\right)=\lambda \left(a{x}^{n+1}+b{x}^{-n}\right) \\ \Rightarrow \lambda =n\left(n+1\right) \end{gathered}$$