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) ((1)/(2))` , then the value of ` alpha ` is equal to _____

Answer

To solve the given differential equation and find the value of \( \alpha \), we will follow these steps:

### Step 1: Write the differential equation
The differential equation is given as:
\[ \frac{dy}{dx} = e^{\alpha x + y} \]

### Step 2: Separate variables
We can rewrite the equation by separating the variables:
\[ \frac{dy}{e^y} = e^{\alpha x} dx \]

### Step 3: Integrate both sides
Now we will integrate both sides:
\[ \int \frac{dy}{e^y} = \int e^{\alpha x} dx \]
The left side integrates to:
\[ -\frac{1}{e^y} = \int e^{\alpha x} dx = \frac{1}{\alpha} e^{\alpha x} + C \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ -\frac{1}{e^y} = \frac{1}{\alpha} e^{\alpha x} + C \]
Multiplying through by \(-1\):
\[ \frac{1}{e^y} = -\frac{1}{\alpha} e^{\alpha x} - C \]
Taking the reciprocal:
\[ e^y = \frac{1}{-\frac{1}{\alpha} e^{\alpha x} - C} \]

### Step 5: Apply initial conditions
We have two initial conditions:
1. \( y(\log 2) = \log 2 \)
2. \( y(0) = \log \frac{1}{2} \)

Using the second condition \( y(0) = \log \frac{1}{2} \):
\[ e^{\log \frac{1}{2}} = \frac{1}{-\frac{1}{\alpha} e^{0} - C} \]
This simplifies to:
\[ \frac{1}{2} = \frac{1}{-\frac{1}{\alpha} - C} \]
Thus:
\[ -\frac{1}{\alpha} - C = 2 \quad \text{(1)} \]

### Step 6: Use the first condition
Now using the first condition \( y(\log 2) = \log 2 \):
\[ e^{\log 2} = \frac{1}{-\frac{1}{\alpha} e^{\alpha \log 2} - C} \]
This simplifies to:
\[ 2 = \frac{1}{-\frac{1}{\alpha} \cdot 2^{\alpha} - C} \]
Thus:
\[ -\frac{1}{\alpha} \cdot 2^{\alpha} - C = \frac{1}{2} \quad \text{(2)} \]

### Step 7: Solve the equations
Now we have two equations:
1. \( -\frac{1}{\alpha} - C = 2 \)
2. \( -\frac{1}{\alpha} \cdot 2^{\alpha} - C = \frac{1}{2} \)

From equation (1):
\[ C = -\frac{1}{\alpha} - 2 \]
Substituting \( C \) into equation (2):
\[ -\frac{1}{\alpha} \cdot 2^{\alpha} + \frac{1}{\alpha} + 2 = \frac{1}{2} \]
This simplifies to:
\[ -\frac{1}{\alpha} \cdot 2^{\alpha} + \frac{1}{\alpha} = \frac{1}{2} - 2 \]
\[ -\frac{1}{\alpha} \cdot 2^{\alpha} + \frac{1}{\alpha} = -\frac{3}{2} \]
Multiplying through by \(-\alpha\):
\[ 2^{\alpha} - 1 = \frac{3\alpha}{2} \]
Rearranging gives:
\[ 2^{\alpha} = \frac{3\alpha}{2} + 1 \]

### Step 8: Find \( \alpha \)
Testing \( \alpha = 2 \):
\[ 2^{2} = 4 \quad \text{and} \quad \frac{3 \cdot 2}{2} + 1 = 3 + 1 = 4 \]
Thus, \( \alpha = 2 \) satisfies the equation.

### Final Answer
The value of \( \alpha \) is:
\[ \boxed{2} \]