Which of the following are solutions of the differential equation $y^{''} - y = 0$ ?

y = C e^{x}

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y = C \frac{e^{x}}{2}

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y = C e^{- x}

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None of these

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Solution

The correct options are
A $y = C e^{x}$
D $y = C e^{- x}$
Given $¨ y - y = 0$
Put $y = e^{m x}$ , we get $m^{2} e^{m x} - e^{m x} = 0$
$\Rightarrow e^{m x} (m^{2} - 1) = 0$
$\Rightarrow m = \pm 1$
Therefore $y = C e^{x}$ and $y = C e^{- x}$ are solutions of given differential equation
Therefore options $A$ and $C$ are correct

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