The curve whose equation satisfies ` x ( dy)/(dx) - 4 y - x^2 sqrt(y) =0=0` passes through `( 1 , (1n 4) ^2)` the find the value of ` ( y(2) )/( ( 1n 32 ) ^(2))`

Answer

To solve the differential equation given by \( x \frac{dy}{dx} - 4y - x^2 \sqrt{y} = 0 \) and find the value of \( \frac{y(2)}{(\ln 32)^2} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x \frac{dy}{dx} - 4y = x^2 \sqrt{y} \]
Dividing through by \( \sqrt{y} \) gives:
\[ \frac{x}{\sqrt{y}} \frac{dy}{dx} - 4\sqrt{y} = x^2 \]

### Step 2: Substitute \( \sqrt{y} \)
Let \( t = \sqrt{y} \). Then, \( y = t^2 \) and \( \frac{dy}{dx} = 2t \frac{dt}{dx} \). Substituting these into the equation gives:
\[ \frac{x}{t} (2t \frac{dt}{dx}) - 4t = x^2 \]
This simplifies to:
\[ 2x \frac{dt}{dx} - 4t = x^2 \]

### Step 3: Rearrange into standard form
Rearranging the equation, we have:
\[ 2x \frac{dt}{dx} = x^2 + 4t \]
Dividing by \( 2x \):
\[ \frac{dt}{dx} = \frac{x}{2} + \frac{2t}{x} \]

### Step 4: Identify as a linear differential equation
This is a linear first-order differential equation in the form:
\[ \frac{dt}{dx} - \frac{2}{x} t = \frac{x}{2} \]

### Step 5: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2} \]

### Step 6: Multiply through by the integrating factor
Multiplying the entire equation by \( \frac{1}{x^2} \):
\[ \frac{1}{x^2} \frac{dt}{dx} - \frac{2}{x^3} t = \frac{1}{2x} \]

### Step 7: Rewrite the left side as a derivative
The left side can be expressed as:
\[ \frac{d}{dx} \left( \frac{t}{x^2} \right) = \frac{1}{2x} \]

### Step 8: Integrate both sides
Integrating both sides gives:
\[ \frac{t}{x^2} = \frac{1}{2} \ln x + C \]

### Step 9: Substitute back for \( t \)
Substituting back \( t = \sqrt{y} \):
\[ \frac{\sqrt{y}}{x^2} = \frac{1}{2} \ln x + C \]
Thus,
\[ \sqrt{y} = x^2 \left( \frac{1}{2} \ln x + C \right) \]
Squaring both sides gives:
\[ y = x^4 \left( \frac{1}{2} \ln x + C \right)^2 \]

### Step 10: Use the initial condition
The curve passes through the point \( (1, (\ln 4)^2) \):
\[ (\ln 4)^2 = 1^4 \left( \frac{1}{2} \ln 1 + C \right)^2 \]
Since \( \ln 1 = 0 \):
\[ (\ln 4)^2 = C^2 \implies C = \ln 4 \]

### Step 11: Substitute \( C \) back into the equation
Thus, we have:
\[ y = x^4 \left( \frac{1}{2} \ln x + \ln 4 \right)^2 \]

### Step 12: Find \( y(2) \)
Substituting \( x = 2 \):
\[ y(2) = 2^4 \left( \frac{1}{2} \ln 2 + \ln 4 \right)^2 = 16 \left( \frac{1}{2} \ln 2 + 2 \ln 2 \right)^2 \]
\[ = 16 \left( \frac{5}{2} \ln 2 \right)^2 = 16 \cdot \frac{25}{4} (\ln 2)^2 = 100 (\ln 2)^2 \]

### Step 13: Compute \( \frac{y(2)}{(\ln 32)^2} \)
Since \( \ln 32 = \ln(2^5) = 5 \ln 2 \):
\[ (\ln 32)^2 = (5 \ln 2)^2 = 25 (\ln 2)^2 \]
Thus,
\[ \frac{y(2)}{(\ln 32)^2} = \frac{100 (\ln 2)^2}{25 (\ln 2)^2} = 4 \]

### Final Answer
The value of \( \frac{y(2)}{(\ln 32)^2} \) is \( \boxed{4} \).