The Solution of $\frac{{dy}}{{dx}} - y = 1,y(0) = 1$ is given by

Given that,
$\frac{{dy}}{{dx}} - y = 1$

$\Rightarrow$ $\frac{{dy}}{{dx}} = 1 + y$

$\Rightarrow$ $\frac{{dy}}{{1 + y}} = dx$

On integrating both sides,

we get
$\log (1 + y) = x + C$
……..(i)
When $x = 0$ and $y = 1$,

then $\log 2 = 0 + c$

$\Rightarrow$ $C = \log 2$

The required solution of the differential equation is$\log (1 + y) = x + \log 2$

$\Rightarrow$ $\log \left( {\frac{{1 + y}}{2}} \right) = x$

$\Rightarrow$ $\frac{{1 + y}}{2} = {e^x}$

$\Rightarrow$ $1 + y = 2{e^x}$

$\Rightarrow$ $y = 2{e^x} - 1$