If $y=\mathrm{A} \cos \mathrm{n} x+\mathrm{B} \sin \mathrm{n} x$, then $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=$

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Aicharya · Accepted Answer

$y=A \cos n x+B \sin n x ...(i)$
Differentiating w.r.to $x$, we get $\begin{aligned} & \frac{d y}{d x}=-A n(\sin n x)+B n \cos n x \\ & \frac{d y}{d x}=-n A \sin n x+n \cdot B \cos n x \end{aligned}$
Again differentiating w.r.to $x$, we get $\begin{aligned} \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2} & =-\mathrm{n}^2 A \cos n x-n^2 B \sin n x \\ & =-n^2[A \cos n x+B \sin n x] \\ \frac{\mathrm{d}^2 y}{d x^2} & =-n^2 y \end{aligned}$ ...[from (i)]

If $y=\mathrm{A} \cos \mathrm{n} x+\mathrm{B} \sin \mathrm{n} x$, then $\frac{\mathrm{d}^2 y}{\mathrm{~d}…

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