` (dy)/(dx) = 3^(x+y), " when " x=y=0`

Answer

To solve the differential equation \(\frac{dy}{dx} = 3^{x+y}\) with the initial condition \(x = 0\) and \(y = 0\), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ \frac{dy}{dx} = 3^{x+y} \]
We can rewrite \(3^{x+y}\) as \(3^x \cdot 3^y\):
\[ \frac{dy}{dx} = 3^x \cdot 3^y \]

### Step 2: Separate Variables
We will separate the variables \(y\) and \(x\). We can rearrange the equation as follows:
\[ \frac{dy}{3^y} = 3^x \, dx \]

### Step 3: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{dy}{3^y} = \int 3^x \, dx \]
The left side can be integrated as:
\[ \int \frac{dy}{3^y} = \frac{-1}{\ln(3)} \cdot 3^{-y} + C_1 \]
The right side integrates to:
\[ \int 3^x \, dx = \frac{3^x}{\ln(3)} + C_2 \]
Thus, we have:
\[ \frac{-1}{\ln(3)} \cdot 3^{-y} = \frac{3^x}{\ln(3)} + C \]
where \(C = C_2 - C_1\).

### Step 4: Simplify the Equation
Multiplying through by \(-\ln(3)\) gives:
\[ 3^{-y} = -3^x - C \ln(3) \]

### Step 5: Solve for \(C\) Using Initial Conditions
Now we apply the initial condition \(x = 0\) and \(y = 0\):
\[ 3^{0} = -3^{0} - C \ln(3) \]
This simplifies to:
\[ 1 = -1 - C \ln(3) \]
Rearranging gives:
\[ C \ln(3) = -2 \quad \Rightarrow \quad C = \frac{-2}{\ln(3)} \]

### Step 6: Substitute \(C\) Back into the Equation
Substituting \(C\) back into the equation gives:
\[ 3^{-y} = -3^x + \frac{2}{\ln(3)} \ln(3) \]
This simplifies to:
\[ 3^{-y} = -3^x + 2 \]

### Step 7: Final Rearrangement
Rearranging gives:
\[ 3^x + 3^{-y} = 2 \]

### Final Answer
The solution to the differential equation is:
\[ 3^x + 3^{-y} = 2 \]