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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
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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