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 \( x \frac{dy}{dx} - 4y - x^2 \sqrt{y} = 0 \) with the initial condition that the curve passes through the point \( (1, (\ln 4)^2) \), we will follow these steps:

### Step-by-step Solution:

1. **Rearranging the Equation:**
Start with the given equation:
\[ x \frac{dy}{dx} - 4y - x^2 \sqrt{y} = 0 \]
Rearranging gives:
\[ x \frac{dy}{dx} = 4y + x^2 \sqrt{y} \]
Dividing both sides by \( x \sqrt{y} \):
\[ \frac{dy}{\sqrt{y}} = \frac{4}{x} + \frac{x}{\sqrt{y}} dx \]

2. **Substituting \( \sqrt{y} = t \):**
Let \( \sqrt{y} = t \), then \( y = t^2 \) and \( dy = 2t dt \). Substituting gives:
\[ \frac{2t dt}{t} = \frac{4}{x} + \frac{x}{t} dx \]
Simplifying:
\[ 2 dt = \frac{4}{x} + \frac{x}{t} dx \]

3. **Rearranging the Equation:**
Rearranging gives:
\[ dt - \frac{2}{x} dt = \frac{2}{x} + \frac{x}{2} \frac{dt}{t} \]
This can be expressed as:
\[ \frac{dt}{dx} + \frac{2}{x} t = \frac{x}{2} \]

4. **Finding the Integrating Factor:**
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int \frac{-2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2} \]

5. **Multiplying through by the Integrating Factor:**
Multiply the entire equation by \( \frac{1}{x^2} \):
\[ \frac{1}{x^2} \frac{dt}{dx} + \frac{2t}{x^3} = \frac{1}{2x} \]

6. **Integrating:**
The left-hand side can be integrated:
\[ \frac{d}{dx}\left(\frac{t}{x^2}\right) = \frac{1}{2x} \]
Integrating both sides gives:
\[ \frac{t}{x^2} = \frac{1}{4} \ln x + C \]

7. **Substituting back for \( t \):**
Recall \( t = \sqrt{y} \):
\[ \frac{\sqrt{y}}{x^2} = \frac{1}{4} \ln x + C \]

8. **Applying the Initial Condition:**
Substitute \( x = 1 \) and \( y = (\ln 4)^2 \):
\[ \frac{\sqrt{(\ln 4)^2}}{1^2} = \frac{1}{4} \ln(1) + C \]
Since \( \ln(1) = 0 \):
\[ \ln 4 = C \]

9. **Final Equation:**
The equation becomes:
\[ \frac{\sqrt{y}}{x^2} = \frac{1}{4} \ln x + \ln 4 \]
Rearranging gives:
\[ \sqrt{y} = x^2 \left(\frac{1}{4} \ln x + \ln 4\right) \]
Squaring both sides gives:
\[ y = x^4 \left(\frac{1}{4} \ln x + \ln 4\right)^2 \]

10. **Finding \( y(2) \):**
Substitute \( x = 2 \):
\[ y(2) = 2^4 \left(\frac{1}{4} \ln 2 + \ln 4\right)^2 = 16 \left(\frac{1}{4} \ln 2 + 2 \ln 2\right)^2 = 16 \left(\frac{9}{4} \ln 2\right)^2 \]
Simplifying gives:
\[ y(2) = 16 \cdot \frac{81}{16} (\ln 2)^2 = 81 (\ln 2)^2 \]

11. **Calculating \( \frac{y(2)}{(\ln 32)^2} \):**
Since \( \ln 32 = 5 \ln 2 \):
\[ \frac{y(2)}{(\ln 32)^2} = \frac{81 (\ln 2)^2}{(5 \ln 2)^2} = \frac{81}{25} = 4 \]

### Final Answer:
\[ \frac{y(2)}{(\ln 32)^2} = 4 \]