The differential equation obtained on eliminating A and B from y=… - Vidyadip

MCQ

The differential equation obtained on eliminating $A$ and $B$ from $\text{y}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}$ is:

A
$\text{y}''+\text{y}'=0$
B
$\text{y}''-\omega^{2}\text{y}=0$
✓
$\text{y}''=-\omega^{2}\text{y}=0$
D
$\text{y}''+\text{y}=0$

✓

Answer

Correct option: C.

$\text{y}''=-\omega^{2}\text{y}=0$

We have,
$\text{y}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}\ ...(\text{i})$
Differentiating both sides of $(i)$ with respect to $x$, we get
$\frac{\text{dy}}{\text{dt}}=-\text{A}\omega\sin\omega\text{t}+\text{B}\omega\cos\omega\text{t}\ ...(\text{ii})$
Differentiating both sides of $(ii)$
$\frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\text{A}\omega^{2}\cos\omega\text{t}+\text{B}\omega^{2}\sin\omega\text{t}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}(\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t})$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}\text{y}$
$\text{y}''=-\omega^{2}\text{y}$

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