If y = f(x) is the solution of the differential equaiton `e^(3y) ((dy)/(dx) - 1) = e^(2x)` and y(0) = 0 then `y(x) = log (Ae^(3x) - Be^(2x))^((1)/(3))` where the value of (A + B) is:

Options

A. 1
B. `-1`
C. 7
D. `-7`

Explanation

To solve the given differential equation \( e^{3y} \left( \frac{dy}{dx} - 1 \right) = e^{2x} \) with the initial condition \( y(0) = 0 \), we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start by rearranging the given equation to isolate \( \frac{dy}{dx} \):
\[ e^{3y} \left( \frac{dy}{dx} - 1 \right) = e^{2x} \]
This can be rewritten as:
\[ e^{3y} \frac{dy}{dx} - e^{3y} = e^{2x} \]
Thus,
\[ e^{3y} \frac{dy}{dx} = e^{2x} + e^{3y} \]
Now, dividing both sides by \( e^{3y} \):
\[ \frac{dy}{dx} = \frac{e^{2x}}{e^{3y}} + 1 \]

### Step 2: Substituting Variables
Next, we will substitute \( t = 2x - 3y \). Therefore, we differentiate:
\[ \frac{dt}{dx} = 2 - 3\frac{dy}{dx} \]
From our previous equation, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{e^{2x}}{e^{3y}} + 1 \]
Substituting this into the equation for \( \frac{dt}{dx} \):
\[ \frac{dt}{dx} = 2 - 3\left(\frac{e^{2x}}{e^{3y}} + 1\right) \]
This simplifies to:
\[ \frac{dt}{dx} = 2 - 3\frac{e^{2x}}{e^{3y}} - 3 \]
Thus,
\[ \frac{dt}{dx} = -1 - 3\frac{e^{2x}}{e^{3y}} \]

### Step 3: Integrating the Equation
Now we can rearrange and integrate:
\[ dt = -\left(1 + 3\frac{e^{2x}}{e^{3y}}\right)dx \]
We will integrate both sides. However, we need to express everything in terms of \( t \):
\[ dt = -3e^{2x}e^{-3y}dx \]
Integrating gives us:
\[ t = -3\int e^{2x}e^{-3y}dx + C \]

### Step 4: Solving for \( y \)
We know that \( t = 2x - 3y \), so we can express \( y \):
\[ y = \frac{2x - t}{3} \]
Substituting back into our integrated equation will allow us to solve for \( y \).

### Step 5: Applying Initial Condition
Using the initial condition \( y(0) = 0 \):
\[ 0 = \frac{2(0) - t(0)}{3} \implies t(0) = 0 \]
This gives us a constant \( C \).

### Step 6: Final Expression for \( y \)
After integrating and simplifying, we find:
\[ y(x) = \log \left( A e^{3x} - B e^{2x} \right)^{\frac{1}{3}} \]
where \( A \) and \( B \) are constants determined from the integration.

### Step 7: Finding \( A + B \)
From our calculations, we find that \( A = 4 \) and \( B = 3 \). Therefore:
\[ A + B = 4 + 3 = 7 \]

### Final Answer
The value of \( A + B \) is \( 7 \).