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

$y = 2e^x - 1$

The correct answer is Option (4) → $y = 2e^x - 1$ ##

Given that, $\frac{dy}{dx} - y = 1$

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

$\Rightarrow \frac{dy}{1 + y} = dx \quad \text{[using variable separable method]}$

On integrating both sides, we get

$\log(1 + y) = x + C \dots(i)$

When $x = 0$ and $y = 1$, then

$\log 2 = 0 + C$

$\Rightarrow C = \log 2$

The required solution is

$\log(1 + y) = x + \log 2 \quad [∵C = \log 2]$

$\log(1 + y) - \log 2 = x$

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

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

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

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