`2e^(x+2y) dx -3dy =0`

Answer

To solve the differential equation \( 2e^{x + 2y} dx - 3dy = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ 2e^{x + 2y} dx = 3 dy \]
Now, we can separate the variables:
\[ \frac{dy}{dx} = \frac{2}{3} e^{x + 2y} \]

### Step 2: Separating Variables
Next, we separate the variables \( y \) and \( x \):
\[ \frac{dy}{e^{2y}} = \frac{2}{3} e^{x} dx \]
This can be rewritten as:
\[ e^{-2y} dy = \frac{2}{3} e^{x} dx \]

### Step 3: Integrating Both Sides
Now we will integrate both sides:
\[ \int e^{-2y} dy = \int \frac{2}{3} e^{x} dx \]
The left side integrates to:
\[ -\frac{1}{2} e^{-2y} \]
The right side integrates to:
\[ \frac{2}{3} e^{x} + C \]
So we have:
\[ -\frac{1}{2} e^{-2y} = \frac{2}{3} e^{x} + C \]

### Step 4: Rearranging the Equation
Now we will rearrange the equation to isolate \( e^{-2y} \):
\[ e^{-2y} = -\frac{4}{3} e^{x} - 2C \]
Let \( C' = -2C \) (where \( C' \) is a new constant):
\[ e^{-2y} = -\frac{4}{3} e^{x} + C' \]

### Step 5: Final Form
Finally, we can express the solution in a more standard form:
\[ 4 e^{x} + 3 e^{-2y} + C = 0 \]
where \( C \) is a constant.

### Summary of the Solution
The solution to the differential equation \( 2e^{x + 2y} dx - 3dy = 0 \) is:
\[ 4 e^{x} + 3 e^{-2y} + C = 0 \]

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