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The differential equation whose solution is $y = {Ae}^{3 x} + {Be}^{- 3 x}$ where A, B are arbitrary constants, is

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$\frac{d^{2} y}{{dx}^{2}} = 9 y$

$\frac{d^{2} y}{{dx}^{2}} = y$

$\frac{d^{2} y}{{dx}^{2}} = 3 y$

$\frac{{dy}^{2}}{{dx}^{2}} = 2 y$

answer is A.

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Detailed Solution

Given $y = A e^{3 x} + B e^{- 3 x} \Rightarrow \frac{d y}{d x} = 3 A e^{3 x} - 3 B e^{- 3 x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 9 A e^{3 x} + 9 B e^{- 3 x} = 9 [A e^{3 x} + B e^{- 3 x}] = 9 y$

The required differential equation is $d^{2} y / {dx}^{2} = 9 y$ .

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