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Solve : $\log \bigg(\large\frac{1}{x}.\frac{dy}{dx}\bigg)$$=2x+5y$

$(a)\;e^{5y}=e^{2x}-x+c \\ (b)\;e^{-5y}=1+2x+x \\ (c)\;e^{-5y}=\frac{5}{2}(e^{2x}-x-2^{2x})+c\\ (d)\;e^{-5y}=5e^{2x}(1-ex)+c $
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$\large\frac{1}{x}.\frac{dy}{dx}$$=e^{2x}.e^{5y}$
$\large\frac{dy}{e^{5y}}$$=x.e^{2x}dx$
$\large\frac{-e^{-5y}}{5}=\frac{x.e^{2x}}{2}-\frac{2^{2x}}{2}$
$e^{-5y}=\large\frac{5}{2} $$(e^{2x}-x.e^{2x})$
Hence c is the correct answer.
answered Feb 7, 2014 by meena.p
 

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