The solution of equation `(2y-1)dx-(2x+3)dy=0` is

Options

A. `(2x-1)/(2y+3)=k`
B. `(2y+1)/(2x-3)=k`
C. `(2x+3)/(2y-1)=k`
D. `(2x-1)/(2y-1)=k`

Explanation

To solve the differential equation \((2y - 1)dx - (2x + 3)dy = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ (2y - 1)dx = (2x + 3)dy \]

### Step 2: Separating Variables
Next, we separate the variables \(x\) and \(y\):
\[ \frac{dx}{2x + 3} = \frac{dy}{2y - 1} \]

### Step 3: Integrating Both Sides
Now, we integrate both sides:
\[ \int \frac{dx}{2x + 3} = \int \frac{dy}{2y - 1} \]

### Step 4: Performing the Integration
Using the integration formula \(\int \frac{dx}{ax + b} = \frac{1}{a} \ln |ax + b| + C\), we perform the integration:
\[ \frac{1}{2} \ln |2x + 3| = \frac{1}{2} \ln |2y - 1| + C \]

### Step 5: Simplifying the Equation
We can simplify the equation by multiplying through by 2:
\[ \ln |2x + 3| = \ln |2y - 1| + 2C \]
Let \(k = e^{2C}\), we can rewrite this as:
\[ \ln |2x + 3| - \ln |2y - 1| = \ln k \]

### Step 6: Using Logarithmic Properties
Using the property of logarithms \(\ln a - \ln b = \ln \frac{a}{b}\):
\[ \ln \left(\frac{2x + 3}{2y - 1}\right) = \ln k \]

### Step 7: Exponentiating Both Sides
Exponentiating both sides gives:
\[ \frac{2x + 3}{2y - 1} = k \]

### Step 8: Final Form of the Solution
Thus, we can express the solution as:
\[ 2x + 3 = k(2y - 1) \]
or equivalently,
\[ \frac{2x + 3}{2y - 1} = k \]

This is the solution of the given differential equation.

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