Question

The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is

Options

• y" + y' = 0

• y" − ω² y = 0

• y" = −ω² y

• y" + y = 0

Answer 1

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

\[\frac{d^2 y}{d t^2} = - A \omega^2 \cos \omega t - B \omega^2 \sin \omega t\]

\[ \Rightarrow \frac{d^2 y}{d t^2} = - \omega^2 \left( A \cos \omega t + B \sin \omega t \right)\]

\[ \Rightarrow \frac{d^2 y}{d t^2} = - \omega^2 y ..........\left[ \text{Using }\left( 1 \right) \right]\]

\[ \therefore y'' = - \omega^2 y\]