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Solve: $ (e^x+e^{-x} )\large\frac{dy}{dx}$$=1- \large\frac{(e^x+e^{-x})^2}{(e^x-e^{-x})}$$y$

$(a)\;y.(e^x-e^{-x})=e^x+e^{-x}+c \\ (b)\;y.(e^x-e^{-x})=\log (e^x+e^{-x})+c \\ (c)\;y.(e^x-e^{-x})=\log (e^x-e^{-x})+c\\ (d)\;y=\log (e^x+e^{-x})+c $

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It is linear differential equation $\large\frac{dy}{dx}+ \bigg( \large\frac{e^x +e^{-x}}{e^x-e^{-x}} \bigg)$$Y=\large\frac{1}{e^x+e^{-x}}$
$I.F =e^{\int \Large\frac{e^x+e^{-x}}{e^x-e^{-x}}dx}$
$e^x-e^{-x}=t$
$(e^x+e^{-x})dx=dt$
$I.F= e ^{\int \large\frac{dt}{t}}$
$\quad= e^{\log t}=t$
$=>e^x-e^{-x}$
$y. (e^x-e^{-x})=\int \large\frac{e^x-e^{-x}}{e^x+e^{-x}} \times dx$
$e^x-e^{-x}=t$
$y. (e^x-e^{-x})= \int \large\frac{dt}{t}$
$y.(e^x-e^{-x})=\log (e^x-e^{-x})+c$
Hence c is the correct answer.
answered Feb 6, 2014 by meena.p
edited Oct 26, 2014 by sharmaaparna1
 

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