1. Given differential equation

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

We need to solve it with the initial condition

$y(0)=0$

and then find $y(1)$.

2. Rewrite the equation

Expand the left side:

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

Now observe that

$\frac{d}{dx}(e^y)=e^y\frac{dy}{dx}$

So the differential equation becomes

$\frac{d}{dx}(e^y)-e^y=e^x$

Let

$u=e^y$

Then the equation reduces to a linear differential equation:

$\frac{du}{dx}-u=e^x$

3. Solve the linear differential equation

We have

$\frac{du}{dx}-u=e^x$

The integrating factor is

$IF=e^{\int -1\,dx}=e^{-x}$

Multiply both sides by $e^{-x}$:

$e^{-x}\frac{du}{dx}-e^{-x}u=1$

The left side is

$\frac{d}{dx}(ue^{-x})=1$

Integrating both sides:

$ue^{-x}=x+C$

So,

$u=e^x(x+C)$

Since $u=e^y$,

$e^y=e^x(x+C)$

Taking natural logarithm,

$y=x+\ln(x+C)$

4. Use the initial condition

Given $y(0)=0$:

$0=0+\ln C$

So,

$\ln C=0 \Rightarrow C=1$

Hence the solution is

$y=x+\ln(x+1)$

5. Find $y(1)$

Substitute $x=1$:

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

6. Compare with options

• A: $2+\ln 2$
• B: $\ln 2$
• C: $1+\ln 2$
• D: $2e$

Thus the correct option is:

$\boxed{\text{C}}$

and

$\boxed{y(1)=1+\ln 2}$