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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$\left[\begin{array}{ccc}1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}1 \\ -1 \\ 1\end{array}\right]$
of linear equations, has infinitely many solutions, then $1+\alpha+\alpha^2=$