Let `=f(x)` be the solution of the diferential equation `y^(')=(3y^(2)+x)/(4y^(2)+5)` where `y(145/4)=0` then which of the following is/are correct

Options

A. `y^(')ge3/4AAx ge 15/4`
B. `int_(15//4)^(27//4)f(x)dx ge 27/8`
C. `int_(15//4)^(27//4)f(x)dx lt 27/8`
D. `f^(')(x) ge 0AAx ge 0`

Explanation

To solve the differential equation \( y' = \frac{3y^2 + x}{4y^2 + 5} \) with the initial condition \( y\left(\frac{145}{4}\right) = 0 \), we will analyze the given equation step by step.

### Step 1: Rewrite the differential equation
The given differential equation is:
\[ y' = \frac{3y^2 + x}{4y^2 + 5} \]

### Step 2: Analyze the expression for \( y' \)
We can express \( y' \) in a more manageable form:
\[ y' = \frac{3y^2}{4y^2 + 5} + \frac{x}{4y^2 + 5} \]

### Step 3: Find the conditions for \( y' \)
To analyze the behavior of \( y' \), we can consider the term \( \frac{x}{4y^2 + 5} \). Since \( 4y^2 + 5 > 0 \) for all \( y \), the sign of \( y' \) is determined by the numerator \( 3y^2 + x \).

### Step 4: Set the condition for \( y' \)
For \( y' \geq \frac{3}{4} \), we need:
\[ \frac{3y^2 + x}{4y^2 + 5} \geq \frac{3}{4} \]
Cross-multiplying gives:
\[ 3y^2 + x \geq \frac{3}{4}(4y^2 + 5) \]
This simplifies to:
\[ 3y^2 + x \geq 3y^2 + \frac{15}{4} \]
Thus, we have:
\[ x \geq \frac{15}{4} \]

### Step 5: Check the initial condition
Given \( y\left(\frac{145}{4}\right) = 0 \), we can substitute \( x = \frac{145}{4} \):
\[ \frac{145}{4} \geq \frac{15}{4} \]
This condition holds true.

### Step 6: Evaluate the integral for option B
We need to evaluate:
\[ \int_{15/4}^{27/4} f(x) \, dx \]
Since \( f(x) \) is above the line defined by \( y = \frac{3}{4}(x - \frac{15}{4}) \), we can find the area under the curve:
- The base of the triangle is \( \frac{27}{4} - \frac{15}{4} = 3 \).
- The height at \( x = \frac{27}{4} \) can be calculated as follows:
\[ y = \frac{3}{4}\left(\frac{27}{4} - \frac{15}{4}\right) = \frac{3}{4} \cdot 3 = \frac{9}{4} \]
So, the area of the triangle is:
\[ \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 3 \cdot \frac{9}{4} = \frac{27}{8} \]
Thus, \( \int_{15/4}^{27/4} f(x) \, dx \geq \frac{27}{8} \).

### Step 7: Check the positivity of \( f'(x) \)
From the expression:
\[ f'(x) = \frac{3}{4} + \frac{x - \frac{15}{4}}{4y^2 + 5} \]
Since \( 4y^2 + 5 > 0 \) and \( x - \frac{15}{4} \) is positive for \( x > \frac{15}{4} \), we conclude that \( f'(x) > 0 \) for \( x > 0 \).

### Conclusion
Based on the analysis:
- Option A is correct: \( y' \geq \frac{3}{4} \) when \( x \geq \frac{15}{4} \).
- Option B is correct: \( \int_{15/4}^{27/4} f(x) \, dx \geq \frac{27}{8} \).
- Option D is correct: \( f'(x) > 0 \) for \( x > 0 \).
- Option C is incorrect.