If `y^(2) = ax^(2) + bx + c` where a, b c are contants then
`y^(3)(d^(2)y)/(dx^(2))=`?

Options

A. A constant
B. A function of x
C. A function of y
D. None of these

Answer

To solve the problem, we need to find the expression for \( y^3 \frac{d^2y}{dx^2} \) given the equation \( y^2 = ax^2 + bx + c \), where \( a, b, c \) are constants.

### Step-by-Step Solution:

1. **Differentiate the given equation:**
\[ y^2 = ax^2 + bx + c \]
Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(ax^2 + bx + c) \]
Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = 2ax + b \]

2. **Rearranging for \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = \frac{2ax + b}{2y} \]

3. **Differentiate again to find \( \frac{d^2y}{dx^2} \):**
We will apply the quotient rule to differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2ax + b}{2y}\right) \]
Let \( u = 2ax + b \) and \( v = 2y \). Then:
\[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where:
\[ \frac{du}{dx} = 2a \]
and:
\[ \frac{dv}{dx} = 2 \frac{dy}{dx} = 2 \cdot \frac{2ax + b}{2y} = \frac{2ax + b}{y} \]
Substituting these into the formula gives:
\[ \frac{d^2y}{dx^2} = \frac{(2y)(2a) - (2ax + b)\left(\frac{2ax + b}{y}\right)}{(2y)^2} \]

4. **Simplifying \( \frac{d^2y}{dx^2} \):**
\[ \frac{d^2y}{dx^2} = \frac{4ay - \frac{(2ax + b)^2}{y}}{4y^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{4ay^2 - (2ax + b)^2}{4y^3} \]

5. **Finding \( y^3 \frac{d^2y}{dx^2} \):**
Multiply \( y^3 \) by \( \frac{d^2y}{dx^2} \):
\[ y^3 \frac{d^2y}{dx^2} = y^3 \cdot \frac{4ay^2 - (2ax + b)^2}{4y^3} \]
This simplifies to:
\[ y^3 \frac{d^2y}{dx^2} = \frac{4ay^2 - (2ax + b)^2}{4} \]

6. **Substituting \( y^2 \):**
Since \( y^2 = ax^2 + bx + c \):
\[ y^3 \frac{d^2y}{dx^2} = \frac{4a(ax^2 + bx + c) - (2ax + b)^2}{4} \]

7. **Final simplification:**
After simplifying the expression, we find that:
\[ y^3 \frac{d^2y}{dx^2} = ac - b^2 \]
This is a constant since \( a, b, c \) are constants.

### Conclusion:
Thus, the final answer is that \( y^3 \frac{d^2y}{dx^2} \) is a constant.