Question

If y = a x^{n + 1} + bx⁻ⁿ and \[x^2 \frac{d^2 y}{d x^2} = \lambda y\]  then write the value of λ ?

Answer 1

\[y = a x^{n + 1} + b x^{- n} \]

\[\text {  and }x^2 \frac{d^2 y}{d x^2} = \lambda y\]

\[\text { Now }, \]

\[\frac{d y}{d x} = a\left( n + 1 \right) x^n - bn x^{- n - 1} \]

\[\text{ and } \frac{d^2 y}{d x^2} = an\left( n + 1 \right) x^{n - 1} - bn\left( - n - 1 \right) x^{- n - 2} \]

\[\text { Now,} x^2 \frac{d^2 y}{d x^2} = \lambda y \left[ \text { Given } \right]\]

\[ \Rightarrow x^2 \left[ an\left( n + 1 \right) x^{n - 1} + bn\left( n + 1 \right) x^{- n - 2} \right] = \lambda\left( a x^{n + 1} + b x^{- n} \right)\]

\[ \Rightarrow an\left( n + 1 \right) x^{n + 1} + bn\left( n + 1 \right) x^{- n} = \lambda\left( a x^{n + 1} + b x^{- n} \right)\]

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

\[ \Rightarrow \lambda = n\left( n + 1 \right)\]