MCQ
If $y = a{x^{n + 1}} + b{x^{ - n}}$, then ${x^2}{{{d^2}y} \over {d{x^2}}} = $
- A$n\,(n - 1)y$
- ✓$n\,(n + 1)y$
- C$ny$
- D${n^2}y$
$\Rightarrow \frac{{dy}}{{dx}} = (n + 1)a{x^n} - nb{x^{ - n - 1}}$
==> $\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)a{x^{n - 1}} + n(n + 1)b{x^{ - n - 2}}$
==> ${x^2}\frac{{{d^2}y}}{{d{x^2}}} = n(n + 1)y$.
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