Form the differential equation for the curve: `y=Ax^2+Bx`, where a and b are arbitrary constant.

Answer

To form the differential equation for the curve given by \( y = Ax^2 + Bx \), where \( A \) and \( B \) are arbitrary constants, we can 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 \( x \):
\[ \frac{dy}{dx} = 2Ax + B \]
Let’s denote the first derivative as \( y' \):
\[ y' = 2Ax + B \]

### Step 2: Differentiate again to find the second derivative
Now, we differentiate \( y' \) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2Ax + B) = 2A \]
Let’s denote the second derivative as \( y'' \):
\[ y'' = 2A \]

### Step 3: Express \( A \) in terms of \( y'' \)
From the second derivative, we can express \( A \):
\[ A = \frac{y''}{2} \]

### Step 4: Substitute \( A \) back into the first derivative
Substituting \( A \) into the equation for \( y' \):
\[ y' = 2\left(\frac{y''}{2}\right)x + B = y''x + B \]

### Step 5: Express \( B \) in terms of \( y' \) and \( y'' \)
From the equation \( y' = y''x + B \), we can express \( B \):
\[ B = y' - y''x \]

### Step 6: Substitute \( A \) and \( B \) back into the original equation
Now we substitute \( A \) and \( B \) back into the original equation \( y = Ax^2 + Bx \):
\[ y = \left(\frac{y''}{2}\right)x^2 + (y' - y''x)x \]
This simplifies to:
\[ y = \frac{y''}{2}x^2 + y'x - y''x^2 \]
Combining like terms, we have:
\[ y = \left(\frac{y''}{2} - y''\right)x^2 + y'x \]
\[ y = -\frac{y''}{2}x^2 + y'x \]

### Step 7: Rearranging to form the differential equation
Rearranging gives us:
\[ \frac{y''}{2}x^2 - y'x + y = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ y''x^2 - 2y' + 2y = 0 \]

### Final Differential Equation
Thus, the differential equation for the curve \( y = Ax^2 + Bx \) is:
\[ x^2y'' - 2y' + 2y = 0 \]