Let y=(x) be the solution curve of the differential equation `dy/dx = y/x (1+xy^(2)(1+log_e x)),x>0,y(1)=3.` Then `(y^2(x))/9` is equal to:

Options

A. `x^2/(5-2x^(3)(2+log_ex^3))`
B. `(x^2)/(3x^(3)(1+log_ex^2)-2)`
C. `x^2/(7-3x^(3)(1+log_ex^2))`
D. `x^2/(2x^(3)(2+log_ex^3)-3)`

Answer

To solve the given differential equation \( \frac{dy}{dx} = \frac{y}{x} (1 + xy^2(1 + \ln x)) \) with the initial condition \( y(1) = 3 \), we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start with the equation:
\[ \frac{dy}{dx} = \frac{y}{x} (1 + xy^2(1 + \ln x)) \]
We can rewrite this as:
\[ \frac{dy}{dx} = \frac{y}{x} + \frac{xy^3(1 + \ln x)}{x} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y}{x} + y^3(1 + \ln x) \]

### Step 2: Separating Variables
We can separate the variables:
\[ \frac{dy}{y^3} = (1 + \ln x) dx - \frac{1}{x} dy \]
Rearranging gives us:
\[ \frac{dy}{y^3} = \left(1 + \ln x - \frac{1}{x}\right) dx \]

### Step 3: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{dy}{y^3} = \int \left(1 + \ln x - \frac{1}{x}\right) dx \]
The left side integrates to:
\[ -\frac{1}{2y^2} \]
The right side can be integrated term by term:
\[ \int 1 \, dx + \int \ln x \, dx - \int \frac{1}{x} \, dx = x + x \ln x - x + C \]
Thus, we have:
\[ -\frac{1}{2y^2} = x \ln x + C \]

### Step 4: Applying the Initial Condition
Using the initial condition \( y(1) = 3 \):
\[ -\frac{1}{2(3^2)} = 1 \cdot \ln(1) + C \]
This simplifies to:
\[ -\frac{1}{18} = 0 + C \implies C = -\frac{1}{18} \]

### Step 5: Substituting Back
Substituting \( C \) back into our equation gives:
\[ -\frac{1}{2y^2} = x \ln x - \frac{1}{18} \]
Multiplying through by -2:
\[ \frac{1}{y^2} = -2x \ln x + \frac{1}{9} \]

### Step 6: Finding \( \frac{y^2}{9} \)
Rearranging gives:
\[ y^2 = \frac{9}{1 - 18x \ln x} \]
Thus, we find:
\[ \frac{y^2}{9} = \frac{1}{1 - 18x \ln x} \]

### Final Answer
The value of \( \frac{y^2(x)}{9} \) is:
\[ \frac{y^2(x)}{9} = \frac{1}{1 - 18x \ln x} \]