We are given the differential equation $dy = e^{\alpha x + y}\,dx,$ where $\alpha \in \mathbb{N}$, with conditions $y(\ln 2)=\ln 2, \qquad y(0)=\ln\left(\frac12\right)=-\ln 2.$

We need to find $\alpha$.

1. Rewrite the differential equation

From $dy = e^{\alpha x + y}\,dx,$ we get $\frac{dy}{dx}=e^{\alpha x+y}=e^{\alpha x}e^y.$

This is separable: $e^{-y}\,dy=e^{\alpha x}\,dx.$

2. Integrate both sides

Integrating, $\int e^{-y}\,dy = \int e^{\alpha x}\,dx.$

Now, $\int e^{-y}\,dy = -e^{-y},$ and for $\alpha\neq 0$, $\int e^{\alpha x}\,dx = \frac{1}{\alpha}e^{\alpha x}+C.$

So, $-e^{-y}=\frac{1}{\alpha}e^{\alpha x}+C.$

Let us write it as $e^{-y}=C-\frac{1}{\alpha}e^{\alpha x}.$

3. Use the given conditions

At $x=0$

Given $y(0)=\ln\left(\frac12\right)=-\ln 2.$ Then $e^{-y(0)}=e^{\ln 2}=2.$

Substitute into the solution: $2=C-\frac{1}{\alpha}.$ So, $C=2+\frac{1}{\alpha}.$

At $x=\ln 2$

Given $y(\ln 2)=\ln 2.$ Then $e^{-y(\ln 2)}=e^{-\ln 2}=\frac12.$

Also, $e^{\alpha x}=e^{\alpha \ln 2}=2^\alpha.$

Substitute: $\frac12=C-\frac{1}{\alpha}2^\alpha.$ Now use $C=2+\frac{1}{\alpha}$: $\frac12=2+\frac{1}{\alpha}-\frac{2^\alpha}{\alpha}.$

Multiply through by $\alpha$: $\frac{\alpha}{2}=2\alpha+1-2^\alpha.$ Rearrange: $2^\alpha=1+2\alpha-\frac{\alpha}{2}=1+\frac{3\alpha}{2}.$ So, $2^{\alpha+1}=2+3\alpha.$

4. Solve for natural number $\alpha$

Check natural numbers:

• For $\alpha=1$: $2^{2}=4, \qquad 2+3(1)=5$ Not equal.

• For $\alpha=2$: $2^{3}=8, \qquad 2+3(2)=8$ Equal.

• For $\alpha=3$: $2^{4}=16, \qquad 2+9=11$ Not equal.

For larger $\alpha$, left side grows much faster, so no further solution.

Hence, $\boxed{\alpha=2}.$

5. Comparison with stored answer

Stored correct answer: $2$

Our derived answer is also $2$, so they agree.