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Step-by-Step Solution
Step 1: Express the given differential equation clearly
We have
$$
\frac{dy}{dx} \;=\; \frac{2^x \, y \;+\; 2^y \cdot 2^x}{2^x \;+\; 2^{\,x+y}\,\ln 2}.
$$
Step 2: Simplify the differential equation
Factor out $2^x$ from the numerator and denominator to get:
$$
\frac{dy}{dx}
\;=\;
\frac{2^x\,(y + 2^y)}{2^x \bigl(1 + 2^y\,\ln 2 \bigr)}
\;=\;
\frac{y + 2^y}{1 + 2^y \,\ln 2}.
$$
(Here, we have divided both numerator and denominator by $2^x$.)
Step 3: Separate the variables
Rearrange to isolate all $y$ terms on one side and $dx$ on the other side:
$$
\frac{dy}{dx}
\;=\;
\frac{y + 2^y}{1 + 2^y \,\ln 2}.
$$
Hence:
$$
\frac{(1 + 2^y)\,\ln 2}{y + 2^y}\, dy
\;=\;
dx.
$$
Step 4: Integrate both sides
Integrate the left-hand side with respect to $y$ and the right-hand side with respect to $x$:
$$
\int \frac{(1 + 2^y)\,\ln 2}{y + 2^y} \, dy
\;=\;
\int 1 \, dx.
$$
Observe that the integrand on the left can be viewed as the derivative of
$\ln\!\bigl|\,y + 2^y\bigr|$
because
$$
\frac{d}{dy}\,\Bigl(\ln\bigl(y + 2^y\bigr)\Bigr)
=
\frac{1 + 2^y \ln 2}{y + 2^y}.
$$
Thus:
$$
\ln\!\bigl|\,y + 2^y\bigr|
\;=\;
x + C,
$$
where $C$ is the constant of integration.
Step 5: Apply the initial condition
We are given $y(0)=0$. Substitute $x = 0$ and $y = 0$ into
$\ln\!\bigl|\,y + 2^y\bigr| = x + C$ to find $C$:
$$
\ln\!\bigl(0 + 2^0\bigr) = \ln(1) = 0,
$$
so
$$
0 = 0 + C
\;\;\Longrightarrow\;\;
C = 0.
$$
Therefore, the relationship simplifies to:
$$
\ln\!\bigl|\,y + 2^y\bigr|
\;=\;
x.
$$
Or equivalently:
$$
x = \ln\bigl(y + 2^y\bigr).
$$
Step 6: Find $x$ when $y=1$
Substitute $y = 1$ into
$x = \ln\bigl(y + 2^y\bigr)$:
$$
x
\;=\;
\ln\!\bigl(1 + 2^1\bigr)
\;=\;
\ln(3).
$$
Since $1 < \ln(3) < 2$, we conclude that for $y=1$, the value of $x$ lies in the interval $(1, 2)$.
Final Answer
The value of $x$ for $y=1$ lies in the interval
$(1, 2)$.
