Write the general solution of the following differential equations
`(dy)/(dx)= (1-2y)/(3x+1)`

Answer

To solve the differential equation

\[ \frac{dy}{dx} = \frac{1 - 2y}{3x + 1}, \]

we will separate the variables and integrate both sides.

### Step 1: Separate the variables

We start by rearranging the equation to separate the variables \(y\) and \(x\):

\[ \frac{dy}{1 - 2y} = \frac{dx}{3x + 1}. \]

### Step 2: Integrate both sides

Now, we will integrate both sides.

**Left side:**

The integral of \(\frac{1}{1 - 2y}\) with respect to \(y\) is:

\[ \int \frac{dy}{1 - 2y} = -\frac{1}{2} \ln |1 - 2y| + C_1, \]

where \(C_1\) is a constant of integration.

**Right side:**

The integral of \(\frac{1}{3x + 1}\) with respect to \(x\) is:

\[ \int \frac{dx}{3x + 1} = \frac{1}{3} \ln |3x + 1| + C_2, \]

where \(C_2\) is another constant of integration.

### Step 3: Combine the results

Now we can set the two integrals equal to each other:

\[ -\frac{1}{2} \ln |1 - 2y| = \frac{1}{3} \ln |3x + 1| + C, \]

where \(C = C_2 - C_1\) is a new constant.

### Step 4: Simplify the equation

To simplify, we can multiply both sides by -2:

\[ \ln |1 - 2y| = -\frac{2}{3} \ln |3x + 1| - 2C. \]

Let \(C' = e^{-2C}\), then we can rewrite the equation as:

\[ |1 - 2y| = C' |3x + 1|^{-\frac{2}{3}}. \]

### Step 5: Remove the absolute value

Assuming \(1 - 2y\) is positive (we can consider the case where it is negative separately), we have:

\[ 1 - 2y = C' (3x + 1)^{-\frac{2}{3}}. \]

### Step 6: Solve for \(y\)

Now, we can solve for \(y\):

\[ 2y = 1 - C' (3x + 1)^{-\frac{2}{3}}, \]
\[ y = \frac{1}{2} - \frac{C'}{2} (3x + 1)^{-\frac{2}{3}}. \]

### General Solution

Thus, the general solution of the differential equation is:

\[ y = \frac{1}{2} - \frac{C}{2} (3x + 1)^{-\frac{2}{3}}, \]

where \(C\) is an arbitrary constant.