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The differential equation obtained by eliminating $a$ and $b$ from $y=ae^{3x}+be^{-3x}$ is

\[\begin{array}{1 1}(1)\frac{d^{2}y}{dx^{2}}+ay=0&(2)\frac{d^{2}y}{dx^{2}}-9y=0\\(3)\frac{d^{2}y}{dx^{2}}-9\frac{dy}{dx}=0&(4)\frac{d^{2}y}{dx^{2}}+9x=0\end{array}\]

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$y=ae^{3x}+be^{-3x}$
$\large\frac{dy}{dx} $$=ae^{3x} \times 3 +be^{-3x} \times -3$
$\qquad= 3ae^{3x}-3be^{-3x}$
$\large\frac{d^2y}{dx^2} $$=3ae^{3x} (3) -3be^{-3x} (-3)$
$\qquad= 9ae^{3x}+9be^{-3x}$
$\large\frac{d^2y}{dx^2}$$=9y$
Hence 2 is the correct answer.
answered May 22, 2014 by meena.p
 

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