Solution of the equation `xdy – [y + xy^3 (1 + log x)] dx = 0` is :

Options

A. `-(x^2)/(y^2)=(2x^3)/(3)(2/3 )+C`
B. `(x^2)/(y^2)=(2x^3)/(3) (2/3 + log x)+C`
C. `=(x^2 )/(y^2)=(x^3)/(3)(2/3 + log x )+c`
D. none of these

Answer

To solve the differential equation \( x dy - [y + xy^3 (1 + \log x)] dx = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ x dy - [y + xy^3 (1 + \log x)] dx = 0 \]
Rearranging gives us:
\[ x dy = [y + xy^3 (1 + \log x)] dx \]

### Step 2: Dividing by \( y^2 \)
Next, we can divide both sides by \( y^2 \):
\[ \frac{x dy}{y^2} = \left( \frac{y}{y^2} + \frac{xy^3 (1 + \log x)}{y^2} \right) dx \]
This simplifies to:
\[ \frac{x dy}{y^2} = \left( \frac{1}{y} + xy (1 + \log x) \right) dx \]

### Step 3: Integrating Both Sides
Now we integrate both sides. The left-hand side becomes:
\[ \int \frac{x dy}{y^2} = -\frac{x}{y} + C_1 \]
For the right-hand side, we need to integrate:
\[ \int \left( \frac{1}{y} + xy (1 + \log x) \right) dx \]
This can be split into two integrals:
\[ \int \frac{1}{y} dx + \int xy (1 + \log x) dx \]

### Step 4: Solving the Right-Hand Side
1. The first integral \( \int \frac{1}{y} dx \) gives:
\[ \frac{x}{y} \]
2. The second integral can be solved using integration by parts. Let:
- \( u = \log x \) and \( dv = x dx \)
- Then \( du = \frac{1}{x} dx \) and \( v = \frac{x^2}{2} \)

Using integration by parts:
\[ \int xy (1 + \log x) dx = \int xy dx + \int xy \log x dx \]
The first part integrates to:
\[ \frac{y x^2}{2} \]
The second part can be solved using integration by parts again.

### Step 5: Combining Results
After integrating both sides and simplifying, we will arrive at a relationship between \( x \) and \( y \).

### Final Solution
The final solution will be in the form:
\[ \frac{x}{y} = \frac{y x^2}{2} + y \log x \cdot \frac{x^2}{2} - \frac{y}{4} x^2 + C \]