Choose the correct answer from the given four option.If y=e^ -x (A +B ), then y is a solution of: 1. d^2y dx^2 +2 dy dx =0 2. d^2y dx^2 -2 dy dx +2y=0 3. d^2y dx^2 +2 dy dx +2y=0 4. d^2y dx^2 +2y=0

3. d^2y dx^2 +2 dy dx +2y=0 Solution: Given that, y=e^ -x (A +B ) Differentiating both sides w.r.t.x, we get dy dx =e^ -x (A +B )+e^ -x (-A +B ) dy dx =-y+e^ -x (-A +B ) Again, Differentiating both sides W.r.t.x, we get d^2y dx^2 =- dy dx +e^ -x (-A -B )-e^ -x (-A +B ) d^2y dx^2 =- dy dx -y- [ dy dx +y ] d^2y dx^2 =-2 dy dx -2y d^2y dx^2 +2 dy dx +2y=0

Choose the correct answer from the given four option.If y=e^ -x (…

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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:

$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\text{y}=0$

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$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$

Solution:
Given that, $\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x})$
Differentiating both sides 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 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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