Let $$y = y(x)$$ be the solution of the differential equation $$dy = e^{\alpha x + y} dx$$; $$\alpha \in N$$. If $$y(\log_e 2) = \log_e 2$$ and $$y(0) = \log_e\left(\frac{1}{2}\right)$$, then the value of $$\alpha$$ is equal to _________.

We start with the differential equation

$$dy = e^{\alpha x + y}\,dx.$$

Dividing both sides by $$dx$$ gives

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

Now separate the variables. We write the right‐hand side as a product:

$$\frac{dy}{dx} = e^{\alpha x}\,e^{y}.$$

Moving all terms in $$y$$ to the left and all terms in $$x$$ to the right gives

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

Next we integrate both sides. For the left side we recall the basic integral

$$\int e^{-y}\,dy = -e^{-y} + C_1,$$

and for the right side we recall

$$\int e^{\alpha x}\,dx = \frac{1}{\alpha}e^{\alpha x} + C_2,$$

where $$C_1$$ and $$C_2$$ are constants of integration. Absorbing the two constants into one overall constant $$C$$, we have

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

Multiplying through by $$-1$$ simplifies the expression:

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

Renaming the constant $$C'$$ as $$C$$ for convenience, we obtain the general implicit solution

$$e^{-y} = C - \frac{1}{\alpha}e^{\alpha x}. \quad -(1)$$

We now use the two given initial conditions to determine the constant $$C$$ and then find $$\alpha$$.

First condition  $$y(0)=\ln\!\left(\dfrac12\right).$$

At $$x=0$$ we have $$y = \ln\!\left(\dfrac12\right) = -\ln 2.$$ Thus

$$e^{-y} = e^{-\bigl(-\ln 2\bigr)} = e^{\ln 2} = 2,$$

and

$$e^{\alpha x} = e^{\alpha\cdot 0} = 1.$$

Substituting these values into (1) gives

$$2 = C - \frac{1}{\alpha}\cdot 1,$$

so that

$$C = 2 + \frac{1}{\alpha}. \quad -(2)$$

Second condition  $$y(\ln 2)=\ln 2.$$

At $$x = \ln 2$$ we have $$y = \ln 2,$$ hence

$$e^{-y} = e^{-\ln 2} = \frac12,$$

and

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

Substituting into (1) and using the value of $$C$$ from (2) we get

$$\frac12 = \left(2 + \frac{1}{\alpha}\right) - \frac{1}{\alpha}2^{\alpha}.$$

Simplifying step by step:

$$\frac12 = 2 + \frac{1}{\alpha} - \frac{2^{\alpha}}{\alpha},$$

$$\frac12 - 2 = \frac{1 - 2^{\alpha}}{\alpha},$$

$$-\frac32 = \frac{1 - 2^{\alpha}}{\alpha}.$$

Multiplying both sides by $$\alpha$$ yields

$$-\frac{3\alpha}{2} = 1 - 2^{\alpha},$$

which can be rearranged as

$$2^{\alpha} - 1 = \frac{3\alpha}{2}. \quad -(3)$$

Because $$\alpha$$ is a natural number, we test successive integers in equation (3).

For $$\alpha = 1$$: the left side is $$2^{1}-1 = 1,$$ while the right side is $$\frac{3\cdot1}{2} = 1.5,$$ which is not equal.

For $$\alpha = 2$$: the left side is $$2^{2}-1 = 3,$$ while the right side is $$\frac{3\cdot2}{2} = 3,$$ which matches exactly.

Hence the natural number $$\alpha$$ satisfying all the given conditions is $$\alpha = 2.$$

So, the answer is $$2$$.