Let `y(x)` be a function satisfying `(d^(2)y)/(dx^(2))-(dy)/(dx)+e^(2x)=0`, y(0)= and `y^(')(0)=1`. If maximum value of `y(x)` is `y(alpha)`, then integral part of `2alpha` is……………..

Answer

To solve the differential equation given in the problem, we will follow these steps:

### Step 1: Write the differential equation
The differential equation is given as:
\[ \frac{d^2y}{dx^2} - \frac{dy}{dx} + e^{2x} = 0 \]

### Step 2: Substitute \( t = \frac{dy}{dx} \)
Let \( t = \frac{dy}{dx} \). Then, we have:
\[ \frac{d^2y}{dx^2} = \frac{dt}{dx} \]
Substituting this into the equation gives:
\[ \frac{dt}{dx} - t + e^{2x} = 0 \]

### Step 3: Rearranging the equation
Rearranging the equation, we get:
\[ \frac{dt}{dx} = t - e^{2x} \]

### Step 4: Finding the integrating factor
The integrating factor \( \mu(x) \) for the equation \( \frac{dt}{dx} - t = -e^{2x} \) is:
\[ \mu(x) = e^{-\int 1 \, dx} = e^{-x} \]

### Step 5: Multiply through by the integrating factor
Multiplying the entire equation by \( e^{-x} \):
\[ e^{-x} \frac{dt}{dx} - e^{-x} t = -e^{x} \]

### Step 6: Integrate both sides
The left side can be written as:
\[ \frac{d}{dx}(t e^{-x}) = -e^{x} \]
Integrating both sides gives:
\[ t e^{-x} = -e^{x} + C \]
where \( C \) is the constant of integration.

### Step 7: Solve for \( t \)
Thus,
\[ t = -e^{2x} + Ce^{x} \]

### Step 8: Use initial conditions to find \( C \)
From the problem, we know \( y'(0) = 1 \):
\[ t(0) = -e^{0} + C e^{0} = 1 \implies -1 + C = 1 \implies C = 2 \]
So,
\[ t = -e^{2x} + 2e^{x} \]

### Step 9: Integrate \( t \) to find \( y \)
Now, we integrate \( t \) to find \( y \):
\[ \frac{dy}{dx} = -e^{2x} + 2e^{x} \]
Integrating gives:
\[ y = -\frac{1}{2} e^{2x} + 2e^{x} + D \]
Using the initial condition \( y(0) = 2 \):
\[ y(0) = -\frac{1}{2} e^{0} + 2e^{0} + D = 2 \implies -\frac{1}{2} + 2 + D = 2 \implies D = \frac{1}{2} \]
Thus, the function \( y \) is:
\[ y = -\frac{1}{2} e^{2x} + 2e^{x} + \frac{1}{2} \]

### Step 10: Find the maximum value of \( y \)
To find the maximum, we set \( \frac{dy}{dx} = 0 \):
\[ -e^{2\alpha} + 2e^{\alpha} = 0 \implies e^{\alpha}(2 - e^{\alpha}) = 0 \]
This gives \( e^{\alpha} = 2 \) or \( \alpha = \ln(2) \).

### Step 11: Calculate \( 2\alpha \)
Thus,
\[ 2\alpha = 2\ln(2) = \ln(4) \]

### Step 12: Find the integral part of \( 2\alpha \)
The integral part of \( \ln(4) \) is:
\[ \lfloor \ln(4) \rfloor \]
Since \( \ln(4) \) is greater than 1 but less than 2, we have:
\[ \lfloor \ln(4) \rfloor = 1 \]

### Final Answer
The integral part of \( 2\alpha \) is:
\[ \boxed{1} \]