Prove that $y = {a e}^{- 2 x} + {b e}^{x}$ is the solution of differential equation $\frac{d^{2} y}{{d x}^{2}} + \frac{d y}{d x} - 2 y = 0$

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Solution

We have,

$y = a e^{- 2 x} + b e^{x}$

On differentiating and we get,

$\frac{d y}{d x} = - 2 a e^{- 2 x} + b e^{x}$

Again differentiating and we get,

$\frac{d^{2} y}{d x^{2}} = 4 a e^{- 2 x} + b e^{x}$

Now,

L.H.S.

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

$\Rightarrow 4 a e^{- 2 x} + b e^{x} - 2 a e^{- 2 x} + b e^{x} - 2 (a e^{- 2 x} + b e^{x})$

$\Rightarrow 4 a e^{- 2 x} + b e^{x} - 2 a e^{- 2 x} + b e^{x} - 2 a e^{- 2 x} - 2 b e^{x}$

$\Rightarrow 4 a e^{- 2 x} - 4 a e^{- 2 x} + 2 b e^{x} - 2 b e^{x}$

$\Rightarrow 0$

Hence, this is the answer.

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