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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
Choose the correct answer from the given four option.If $\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x}),$ then $y$ is a solution of:
  • A
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$
  • B
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
  • ✓
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
  • D
    $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\text{y}=0$
✓

Answer

Correct option: C.
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
Given that, $\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x})$
Differentiating both sides $\text{w.r.t.x,}$ we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x})+\text{e}^{-\text{x}}(-\text{A}\sin\text{x}+\text{B}\cos\text{x})$
$\frac{\text{dy}}{\text{dx}}=-\text{y}+\text{e}^{-\text{x}}(-\text{A}\sin\text{x}+\text{B}\cos\text{x})$
Again, Differentiating both sides $\text{W.r.t.x,}$ we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{e}^{-\text{x}}(-\text{A}\cos\text{x}-\text{B}\sin\text{x})-\text{e}^{-\text{x}}(-\text{A}\sin\text{x}+\text{B}\cos\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}-\Big[\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-2\frac{\text{d}\text{y}}{\text{d}\text{x}}-2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$

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