` (dy)/(dx) = 4^(x+y)`

Answer

To solve the differential equation \(\frac{dy}{dx} = 4^{x+y}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \frac{dy}{dx} = 4^{x+y} \]
We can express \(4^{x+y}\) as \(4^x \cdot 4^y\):
\[ \frac{dy}{dx} = 4^x \cdot 4^y \]

### Step 2: Separate variables
We can separate the variables \(y\) and \(x\) by rearranging the equation:
\[ \frac{dy}{4^y} = 4^x \, dx \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side requires the integral of \(\frac{1}{4^y}\):
\[ \int \frac{dy}{4^y} = \int 4^x \, dx \]
The integral of \(\frac{1}{4^y}\) can be rewritten as:
\[ \int 4^{-y} \, dy = -\frac{1}{\log 4} \cdot 4^{-y} + C_1 \]
And the integral of \(4^x\) is:
\[ \int 4^x \, dx = \frac{4^x}{\log 4} + C_2 \]
Thus, we have:
\[ -\frac{1}{\log 4} \cdot 4^{-y} = \frac{4^x}{\log 4} + C \]
where \(C\) is a constant that combines \(C_1\) and \(C_2\).

### Step 4: Simplify the equation
Multiplying through by \(-\log 4\) to eliminate the fraction gives:
\[ 4^{-y} = -4^x - C \log 4 \]
We can rewrite this as:
\[ 4^{-y} + 4^x = -C \log 4 \]

### Step 5: Express the solution
Let \(A = -C \log 4\) (where \(A\) is a constant), we can express our solution as:
\[ 4^x + 4^{-y} = A \]

This is the general solution to the differential equation.

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