The differential equation of the family of curves `y=k_(1)x^(2)+k_(2)` is given by (where, `k_(1) and k_(2)` are arbitrary constants and `y_(1)=(dy)/(dx), y_(2)=(d^(2)y)/(dx^(2))`)

Options

A. `y_(1)=x^(2)y_(2)`
B. `(y_(1))^(2)=xy_(2)`
C. `xy_(2)=y_(1)`
D. `y_(1)y_(2)=x`

Answer

To find the differential equation of the family of curves given by \( y = k_1 x^2 + k_2 \), where \( k_1 \) and \( k_2 \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ y = k_1 x^2 + k_2 \]
Now we differentiate this with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(k_1 x^2 + k_2) \]
Since \( k_2 \) is a constant, its derivative is zero. Thus, we have:
\[ \frac{dy}{dx} = k_1 \cdot 2x = 2k_1 x \]

### Step 2: Solve for \( k_1 \)
From the equation obtained in Step 1, we can express \( k_1 \) in terms of \( \frac{dy}{dx} \):
\[ k_1 = \frac{1}{2x} \frac{dy}{dx} \]

### Step 3: Differentiate again to find \( \frac{d^2y}{dx^2} \)
Next, we differentiate \( \frac{dy}{dx} = 2k_1 x \) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = 2k_1 \cdot \frac{d}{dx}(x) + 2x \cdot \frac{dk_1}{dx} \]
Since \( k_1 \) is a constant with respect to \( x \), \( \frac{dk_1}{dx} = 0 \). Therefore:
\[ \frac{d^2y}{dx^2} = 2k_1 \]

### Step 4: Solve for \( k_1 \) again
From the equation obtained in Step 3, we can express \( k_1 \) in terms of \( \frac{d^2y}{dx^2} \):
\[ k_1 = \frac{1}{2} \frac{d^2y}{dx^2} \]

### Step 5: Equate the two expressions for \( k_1 \)
Now we have two expressions for \( k_1 \):
1. \( k_1 = \frac{1}{2x} \frac{dy}{dx} \)
2. \( k_1 = \frac{1}{2} \frac{d^2y}{dx^2} \)

Setting them equal gives:
\[ \frac{1}{2x} \frac{dy}{dx} = \frac{1}{2} \frac{d^2y}{dx^2} \]

### Step 6: Simplify the equation
Multiplying both sides by 2 to eliminate the fractions:
\[ \frac{dy}{dx} = x \frac{d^2y}{dx^2} \]

### Conclusion
Thus, the differential equation of the family of curves is:
\[ \frac{dy}{dx} = x \frac{d^2y}{dx^2} \]