what is the solution of (1 +2x) dy-(1 -2y) dx=0 ?

Options

A. x -y-2xy =c
B. y -x-2xy =c
C. y +x-2xy =c
D. x+y+2xy =c

Explanation

To solve the differential equation \( (1 + 2x) dy - (1 - 2y) dx = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to separate the variables \( y \) and \( x \):
\[ (1 + 2x) dy = (1 - 2y) dx \]
Now, we can divide both sides by \( (1 - 2y)(1 + 2x) \):
\[ \frac{dy}{1 - 2y} = \frac{dx}{1 + 2x} \]

### Step 2: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{dy}{1 - 2y} = \int \frac{dx}{1 + 2x} \]

### Step 3: Finding the Integrals
The left-hand side integral can be solved using the substitution \( u = 1 - 2y \), which gives \( du = -2 dy \) or \( dy = -\frac{1}{2} du \):
\[ \int \frac{dy}{1 - 2y} = -\frac{1}{2} \ln |1 - 2y| + C_1 \]
For the right-hand side, we can use the substitution \( v = 1 + 2x \), which gives \( dv = 2 dx \) or \( dx = \frac{1}{2} dv \):
\[ \int \frac{dx}{1 + 2x} = \frac{1}{2} \ln |1 + 2x| + C_2 \]

### Step 4: Setting the Integrals Equal
Now we set the two integrals equal to each other:
\[ -\frac{1}{2} \ln |1 - 2y| = \frac{1}{2} \ln |1 + 2x| + C \]
where \( C = C_2 - C_1 \).

### Step 5: Simplifying the Equation
Multiplying through by -2 gives:
\[ \ln |1 - 2y| = -\ln |1 + 2x| - 2C \]
Using properties of logarithms, we can rewrite this as:
\[ \ln |1 - 2y| + \ln |1 + 2x| = -2C \]
This can be simplified to:
\[ \ln |(1 - 2y)(1 + 2x)| = -2C \]

### Step 6: Exponentiating Both Sides
Exponentiating both sides gives:
\[ |(1 - 2y)(1 + 2x)| = e^{-2C} \]
Let \( K = e^{-2C} \), then we can write:
\[ (1 - 2y)(1 + 2x) = K \]

### Step 7: Final Form of the Solution
Rearranging gives us the general solution:
\[ 1 - 2y - 2xy = K \]
or equivalently,
\[ x - y - 2xy = C \]
where \( C \) is a constant.

### Final Answer
Thus, the solution to the differential equation is:
\[ x - y - 2xy = C \]