What is the differential equation of the curve ` y=ax^(2)+bx` ?

Options

A. `x^(2)(d^(2)y)/(dx^(2))-2x (dy)/(dx)+2y=0`
B. `x^(2) (d^(2)y)/(dx^(2))-y((dy)/(dx))^(2)+2=0`
C. ` (1-x^(2))(d^(2)y)/(dx^(2)) =(y(dy)/(dx))^(2) =0`
D. none of the above

Explanation

To find the differential equation of the curve given by \( y = ax^2 + bx \), we will follow these steps:

### Step 1: Differentiate the equation with respect to \( x \)
We start with the equation:
\[ y = ax^2 + bx \]
Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = 2ax + b \]
This is our first equation.

### Step 2: Differentiate again to find the second derivative
Now, we differentiate \( \frac{dy}{dx} \) again with respect to \( x \):
\[ \frac{d^2y}{dx^2} = 2a \]
This gives us the second derivative.

### Step 3: Express \( a \) in terms of the second derivative
From the equation \( \frac{d^2y}{dx^2} = 2a \), we can express \( a \) as:
\[ a = \frac{1}{2} \frac{d^2y}{dx^2} \]

### Step 4: Substitute \( a \) back into the first derivative equation
Now, we substitute \( a \) back into the first derivative equation:
\[ \frac{dy}{dx} = 2\left(\frac{1}{2} \frac{d^2y}{dx^2}\right)x + b \]
This simplifies to:
\[ \frac{dy}{dx} = x \frac{d^2y}{dx^2} + b \]

### Step 5: Express \( b \) in terms of the first derivative and the second derivative
Rearranging the equation gives us:
\[ b = \frac{dy}{dx} - x \frac{d^2y}{dx^2} \]

### Step 6: Substitute \( a \) and \( b \) back into the original equation
Now we substitute \( a \) and \( b \) back into the original equation \( y = ax^2 + bx \):
\[ y = \left(\frac{1}{2} \frac{d^2y}{dx^2}\right)x^2 + \left(\frac{dy}{dx} - x \frac{d^2y}{dx^2}\right)x \]
This simplifies to:
\[ y = \frac{1}{2} \frac{d^2y}{dx^2} x^2 + x \frac{dy}{dx} - x^2 \frac{d^2y}{dx^2} \]

### Step 7: Rearranging the equation
Combining like terms gives:
\[ y = \left(\frac{1}{2} - 1\right)x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} \]
This simplifies to:
\[ y = -\frac{1}{2} x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} \]

### Step 8: Form the differential equation
Rearranging gives us:
\[ x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0 \]

Thus, the differential equation of the curve \( y = ax^2 + bx \) is:
\[ x^2 \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + 2y = 0 \]