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प्रश्न
If y = axn+1 + bx−n, then \[x^2 \frac{d^2 y}{d x^2} =\]
पर्याय
n (n − 1)y
n (n − 1)y
ny
n2y
उत्तर
(b) n (n+1)y
Here,
\[y = a x^{n + 1} + b x^{- n} \]
\[ \Rightarrow \frac{d y}{d x} = a\left( n + 1 \right) x^n - bn x^{- n - 1} \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = an\left( n + 1 \right) x^{n - 1} + bn\left( n + 1 \right) x^{- n - 2} \]
\[ \therefore x^2 \frac{d^2 y}{d x^2} = x^2 \left\{ an\left( n + 1 \right) x^{n - 1} + bn\left( n + 1 \right) x^{- n - 2} \right\}\]
\[ = n\left( n + 1 \right)\left( a x^{n + 1} + b x^{- n} \right)\]
\[ = n\left( n + 1 \right)y\]
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