If $y=\mathrm{ax}^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}$, then $x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}=$

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Aicharya · Accepted Answer

$y=a x^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}$
Differentiating w.r.t. $x$, we get $\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{a}(\mathrm{n}+1) x^{\mathrm{n}+1-1}+\mathrm{b}(-\mathrm{n}) x^{-\mathrm{n}-1} \\ & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\mathrm{a}(\mathrm{n}+1) x^{\mathrm{n}}-\mathrm{bn} x^{-\mathrm{n}-1} \end{aligned}$
Again differentiating w.r.t. $x$, we get $\begin{aligned} \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} & =a(n+1)^n x^{\mathrm{n}-1}-b n(-n-1) x^{-n-1-1} \\ & =a(n+1) n x^{n-1}+b n(n+1) x^{-n-2} \\ & =n(n+1) \frac{a x^n}{x}+n(n+1) b \frac{x^{-n}}{x^2} \end{aligned}$ $\begin{array}{ll} & \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{n}(\mathrm{n}+1)}{x^2}\left[\mathrm{a} x^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}\right] \\ \therefore \quad & x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}=\mathrm{n}(\mathrm{n}+1) y\end{array}$

If $y=\mathrm{ax}^{\mathrm{n}+1}+\mathrm{b} x^{-\mathrm{n}}$, then $x^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d}…

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