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

Options

A. `x^(2) "" (d^(2) y)/(dx^(2)) - 2 x"" (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 these

Answer

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

### Step 1: Differentiate the given equation
We start with the equation of the curve:
\[ y = ax^2 + bx \]
We differentiate both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(ax^2) + \frac{d}{dx}(bx) \]
Using the power rule for differentiation, we get:
\[ \frac{dy}{dx} = 2ax + b \]

### Step 2: Differentiate again to find the second derivative
Now we differentiate \( \frac{dy}{dx} \) again with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2ax + b) \]
Since \( a \) and \( b \) are constants, we have:
\[ \frac{d^2y}{dx^2} = 2a \]

### 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
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 \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \)
From the equation above, we can isolate \( b \):
\[ b = \frac{dy}{dx} - x \frac{d^2y}{dx^2} \]

### 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:
\[ y = \frac{1}{2} \frac{d^2y}{dx^2} x^2 + \frac{dy}{dx} x - x^2 \frac{d^2y}{dx^2} \]

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

### Step 8: Rearranging to form the differential equation
Rearranging this equation leads us to the final form of the differential equation:
\[ x^2 \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0 \]

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