Question

The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
(a) y" + y' = 0
(b) y" − ω² y = 0
(c) y" = −ω² y
(d) y" + y = 0

Answer 1

(c) y" = −ω² y

We have,
y = A cos ωt + B sin ωt .....(1)
Differentiating both sides of (1) with respect to x, we get
$\frac{dy}{dt}=-A\omega \sin \omega t+B\omega \cos \omega t$ .....(2)
Differentiating both sides of (2) again with respect to x, we get
$$\begin{gathered} \frac{d^{2}y}{d{t}^2}=-A{\omega }^2\cos \omega t-B{\omega }^2\sin \omega t \\ \Rightarrow \frac{d^{2}y}{d{t}^2}=-{\omega }^2\left(A\cos \omega t+B\sin \omega t\right) \\ \Rightarrow \frac{d^{2}y}{d{t}^2}=-{\omega }^{2}y\left[\text{Using}\left(1\right)\right] \\ \therefore y\text{'}\text{'}=-{\omega }^{2}y \end{gathered}$$