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Solution of $\large\frac{dx}{dy}$$+mx=0,$where $m<0$ is

\[\begin{array}{1 1}(1)x=ce^{my}&(2)x=ce^{-my}\\(3)x=my+c&(4)x=c\end{array}\]

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$\large\frac{dx}{dy}$$+mx=0$
$\large\frac{dx}{dy}$$=-mx$
$\large\frac{dx}{x}$$=-m \;dy$
$\log x =-my + \log c$
$\log x - \log C =-my$
$\log \large\frac{x}{c}$$ =-my$
$\large\frac{x}{c} =e^{-my}$
$x= ce^{-my}$
Hence 2 is the correct answer.
answered May 21, 2014 by meena.p
 

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