MCQ
Which of the following differentials equation has $\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$ as the general solution?
  • A
    $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
  • $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
  • C
    $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{1}=0$
  • D
    $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{1}=0$

Answer

Correct option: B.
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
We have,
$\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{i})$
Differentiating both sides of $(i)$ with we get,
$\frac{\text{dy}}{\text{dx}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{ii})$
Differentiating both sides of $(ii)$ with we get,
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{y}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$

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