MCQ
જો $y = a{x^{n + 1}} + b{x^{ - n}},$ તો ${x^2}\frac{{{d^2}y}}{{d{x^2}}}$ મેળવો.
- A$n(n - 1)y$
- ✓$n(n + 1)y$
- C$ny$
- D$n^2y$
Differentiate with respect to $x$ , $\frac{{dy}}{{dx}} = a(n + 1){x^n} - bn{x^{ - n - 1}}$
Again differentiate, $\frac{{{d^2}y}}{{d{x^2}}} = a\,n\,(n + 1){x^{n - 1}} + b\,n\,(n + 1){x^{ - n - 2}}$
==> ${x^2}\frac{{{d^2}y}}{{d{x^2}}} = a\,n\,(n + 1){x^{n + 1}} + b\,n\,(n + 1){x^{ - n}}$
==> ${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)y$.
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