Solve the following differential equations :
` y -x (dy)/(dx) =3 (1+x^(2) (dy)/(dx))`

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y - x \frac{dy}{dx} = 3 + 3x^2 \frac{dy}{dx} \]
Rearranging gives:
\[ y = 3 + (3x^2 + x) \frac{dy}{dx} \]

### Step 2: Isolating \(\frac{dy}{dx}\)
We can isolate \(\frac{dy}{dx}\):
\[ y - 3 = (3x^2 + x) \frac{dy}{dx} \]
Thus,
\[ \frac{dy}{dx} = \frac{y - 3}{3x^2 + x} \]

### Step 3: Separating Variables
We can separate the variables:
\[ \frac{dy}{y - 3} = \frac{dx}{3x^2 + x} \]

### Step 4: Integrating Both Sides
Now we will integrate both sides:
\[ \int \frac{dy}{y - 3} = \int \frac{dx}{3x^2 + x} \]

The left-hand side integrates to:
\[ \ln |y - 3| \]

For the right-hand side, we can factor the denominator:
\[ 3x^2 + x = x(3x + 1) \]
We will use partial fraction decomposition:
\[ \frac{1}{x(3x + 1)} = \frac{A}{x} + \frac{B}{3x + 1} \]
Multiplying through by the denominator:
\[ 1 = A(3x + 1) + Bx \]
Setting \(x = 0\) gives \(A = 1\). Setting \(x = -\frac{1}{3}\) gives \(B = -3\). Thus,
\[ \frac{1}{x(3x + 1)} = \frac{1}{x} - \frac{3}{3x + 1} \]

Now we can integrate:
\[ \int \left( \frac{1}{x} - \frac{3}{3x + 1} \right) dx = \ln |x| - \ln |3x + 1| + C \]

### Step 5: Combining Results
Combining both integrals, we have:
\[ \ln |y - 3| = \ln |x| - \ln |3x + 1| + C \]
This can be simplified to:
\[ \ln |y - 3| = \ln \left( \frac{x}{3x + 1} \right) + C \]

### Step 6: Exponentiating Both Sides
Exponentiating both sides gives:
\[ |y - 3| = K \frac{x}{3x + 1} \]
where \(K = e^C\).

### Step 7: Final Solution
Thus, the final solution can be expressed as:
\[ y - 3 = K \frac{x}{3x + 1} \]
or
\[ y = 3 + K \frac{x}{3x + 1} \]