The differential equation obtained by eliminating $\mathrm{A}$ and $\mathrm{B}$ from $y=A \cos \omega t+B…

The differential equation obtained by eliminating $\mathrm{A}$ and $\mathrm{B}$ from $y=A \cos \omega t+B \sin \omega t$
  1. $\frac{d^2 y}{d t^2}+\omega^2 y=0$
  2. $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dt}^2}+\omega \mathrm{y}^2=0$
  3. $\frac{d^2 y}{d t^2}-\omega^2 y=0$
  4. $\frac{d^2 y}{d t^2}-\omega y^2=0$

Solution

$\begin{aligned} & y=A \cos \omega t+B \sin \omega t \\ & \therefore \quad \frac{d y}{d t}=-A \omega \sin \omega t+B \omega \cos \omega t \\ & \frac{d^2 y}{d t^2}=-A \omega^2 \cos \omega t-B \omega^2 \sin \omega t \\ & =-\omega^2(A \cos \omega t+B \sin \omega t)=-\omega^2 y \\ & \therefore \quad \frac{d^2 y}{d t^2}+\omega^2 y=0\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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