Question

If y = axⁿ⁺¹ + bx⁻ⁿ, then \[x^2 \frac{d^2 y}{d x^2} =\] 

 

विकल्प

• n (n − 1)y

• n (n − 1)y

•  ny

•  n²y

Answer 1

(b) n (n+1)y

Here,

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

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

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

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

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

\[ = n\left( n + 1 \right)y\]