Browse Questions

# 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 $y(0)=50$

$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.