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Which one of the following curves represents the solution of the initial value problem $y= D(e^x-1)$ where $D= \large\frac{dy}{dx}$ and $y(0)=50$

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$y= \large\frac{dy}{dx} $$(e^x-1)$
$\int \large\frac{dx}{e^x-1} =\int \large\frac{dy}{y} $
$e^x-1=t$
$e^x dx=dt$
$\int \large\frac{dt}{(1+t)(t)}$$=\log y+c$
$\int \bigg( \large\frac{t+1}{t(t+1)}-\frac{t}{t(t+1)}\bigg) $$dt=\log y +c$
$\log t -\log (t+1)=\log y +c$
$\log \bigg( \large\frac{e^x-1}{e^x}\bigg)$$= \log y+c$
$\log \bigg(\large\frac{e^x-1}{e^x}\bigg) $$=\log c'y$
$\large\frac{e^x-1}{e^x} $$=c'y$
$(1-e^{-x} )=c' (y)$
$y(0)=50$
$1-e^0= c'(50)$
$c'=\large\frac{1}{50}$
$y= 50 (1-e^{-x})$
Hence d is the correct answer.
answered Feb 5, 2014 by meena.p
 

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