If $$y = y(x)$$ is the solution of the differential equation, $$e^y\left(\frac{dy}{dx} - 1\right) = e^x$$ such that $$y(0) = 0$$, then $$y(1)$$ is equal to

We start with the given differential equation

$$e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}.$$

First, divide both sides by $$e^{y}$$ to make the derivative term explicit. This gives

$$\frac{dy}{dx}-1=e^{x-y}.$$

Now add $$1$$ to both sides so that $$\dfrac{dy}{dx}$$ appears alone on the left:

$$\frac{dy}{dx}=1+e^{x-y}.$$

The presence of the combination $$x-y$$ in the exponential suggests the substitution

$$v=x-y.$$

Because $$y=x-v$$, we find its derivative with respect to $$x$$:

$$\frac{dy}{dx}=1-\frac{dv}{dx}.$$

Substitute this expression for $$\dfrac{dy}{dx}$$ back into the equation $$\dfrac{dy}{dx}=1+e^{x-y}$$. We obtain

$$1-\frac{dv}{dx}=1+e^{v},$$

because $$x-y=v$$ implies $$e^{x-y}=e^{v}.$$

Subtract $$1$$ from both sides:

$$-\frac{dv}{dx}=e^{v}.$$

Multiply both sides by $$-1$$ to isolate $$\dfrac{dv}{dx}$$ with a positive coefficient:

$$\frac{dv}{dx}=-e^{v}.$$

We now have a separable first-order differential equation. Rewrite it in the separated form

$$\frac{dv}{e^{v}}=-\,dx.$$

Recall the integral formula $$\displaystyle\int e^{-v}\,dv=-e^{-v}+C$$. Express $$\dfrac{1}{e^{v}}$$ as $$e^{-v}$$ and integrate both sides:

$$\int e^{-v}\,dv=\int -\,dx.$$

The left integral evaluates to $$-e^{-v}$$, while the right integral simply gives $$-x+C$$, where $$C$$ is an arbitrary constant of integration. Hence

$$-e^{-v}=-x+C.$$

Multiply through by $$-1$$ to tidy the constants:

$$e^{-v}=x-C.$$

Rename $$-C$$ as a new constant $$C_1$$ for simplicity:

$$e^{-v}=x+C_1.$$

Take the reciprocal exponential to a logarithmic form by noticing that $$e^{-v}=e^{-(x-y)}=e^{y-x}$$. Therefore

$$e^{y-x}=x+C_1.$$

Apply the natural logarithm on both sides to solve for $$y$$:

$$y-x=\ln\!\bigl(x+C_1\bigr).$$

Add $$x$$ to both sides, giving the general solution

$$y=x+\ln\!\bigl(x+C_1\bigr).$$

Use the initial condition $$y(0)=0$$ to determine the constant $$C_1$$. Substituting $$x=0$$ and $$y=0$$ gives

$$0=0+\ln\!\bigl(0+C_1\bigr)\quad\Longrightarrow\quad \ln C_1=0.$$

The only positive number whose natural logarithm is zero is $$1$$, so

$$C_1=1.$$

Therefore the particular solution satisfying the initial condition is

$$y=x+\ln\!\bigl(x+1\bigr).$$

We now evaluate this at $$x=1$$:

$$y(1)=1+\ln(1+1)=1+\ln 2.$$

Hence, the correct answer is Option A.