If $$\frac{dy}{dx} = \frac{2^{x+y} - 2^x}{2^y}$$, $$y(0) = 1$$, then $$y(1)$$ is equal to:

We are given the differential equation $$\frac{dy}{dx}= \frac{2^{x+y}-2^x}{2^y}$$ with the condition $$y(0)=1$$ and we must find $$y(1)$$.

First we simplify the right-hand side. We have $$2^{x+y}=2^x\cdot2^y$$, so

$$\frac{2^{x+y}-2^x}{2^y}= \frac{2^x\cdot2^y-2^x}{2^y}=2^x\cdot\frac{2^y-1}{2^y}=2^x\left(1-\frac{1}{2^y}\right).$$

Thus the equation becomes $$\frac{dy}{dx}=2^x\Bigl(1-2^{-y}\Bigr).$$

To separate the variables we divide by $$1-2^{-y}$$ and multiply by $$dx$$, obtaining

$$\frac{dy}{1-2^{-y}} = 2^x\,dx.$$

Now we integrate both sides. For the left side we set $$u=2^{-y}$$, so $$\ln u = -y\ln2 \;\Longrightarrow\; y=-\frac{\ln u}{\ln2},$$ hence $$dy=-\frac{1}{\ln2}\frac{du}{u}.$$

Substituting into the integral gives

$$\int\frac{dy}{1-2^{-y}} = -\frac{1}{\ln2}\int\frac{du}{u(1-u)}.$$

We use partial fractions: $$\frac{1}{u(1-u)}=\frac{1}{u}+\frac{1}{1-u},$$ so

$$-\frac{1}{\ln2}\int\left(\frac{1}{u}+\frac{1}{1-u}\right)du = -\frac{1}{\ln2}\Bigl(\ln|u| - \ln|1-u|\Bigr) = \frac{1}{\ln2}\ln\left|\frac{1-u}{u}\right|.$$

Replacing $$u$$ by $$2^{-y}$$ gives

$$\int\frac{dy}{1-2^{-y}} = \frac{1}{\ln2}\ln\Bigl(2^y-1\Bigr).$$

For the right side we use the standard formula $$\int a^x\,dx=\frac{a^x}{\ln a}+C.$$ With $$a=2$$ we get

$$\int 2^x\,dx=\frac{2^x}{\ln2}+C.$$

Equating the two antiderivatives and multiplying through by $$\ln2$$ yields

$$\ln\!\bigl(2^y-1\bigr)=2^x+C_1,$$

where $$C_1$$ is a constant.

We apply the initial condition $$y(0)=1$$. Substituting $$x=0,\;y=1$$ gives

$$\ln\!\bigl(2^1-1\bigr)=\ln 1=0=2^0+C_1=1+C_1,$$

so $$C_1=-1.$$

Thus $$\ln\!\bigl(2^y-1\bigr)=2^x-1.$$

Exponentiating both sides we obtain

$$2^y-1=e^{\,2^x-1}, \qquad\text{so}\qquad 2^y=1+e^{\,2^x-1}.$$

Taking logarithm base 2 finally gives

$$y=\log_2\!\bigl(1+e^{\,2^x-1}\bigr).$$

Now we need $$y(1)$$. Putting $$x=1$$ makes $$2^x=2$$, hence

$$y(1)=\log_2\!\bigl(1+e^{\,2-1}\bigr)=\log_2(1+e).$$

Hence, the correct answer is Option D.