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If $y=ke^{\lambda x}$ then its differential equation is

\[\begin{array}{1 1}(1)\frac{dy}{dx}=\lambda y&(2)\frac{dy}{dx}=ky\\(3)\frac{dy}{dx}+ky=0&(4)\frac{dy}{dx}=e^{\lambda x}\end{array}\]

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$y=ke^{\lambda x}$
$\large\frac{dy}{dx} $$=k.e^{\lambda x} .\lambda$
$\frac{dy}{dx}=\lambda y$
Hence 1 is the correct answer.
answered May 21, 2014 by meena.p
 

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