मराठी

If Y = Axn+1 + Bx−N, Then X 2 D 2 Y D X 2 = (A) N (N − 1)Y (B) N (N + 1)Y (C) Ny (D) N2y - Mathematics

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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

MCQ
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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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पाठ 12: Higher Order Derivatives - Exercise 12.3 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 12 Higher Order Derivatives
Exercise 12.3 | Q 3 | पृष्ठ २३

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