The differential equation corresponding to the family of curves `y=e^x (ax+ b)` is

Options

A. `(d^(2)y)/(dx^(2))+2(dy)/(dx)-y=0`
B. `(d^(2)y)/(dx^(2))-2(dy)/(dx)-y=0`
C. `(d^(2)y)/(dx^(2))+2(dy)/(dx)+y=0`
D. `(d^(2)y)/(dx^(2))-2(dy)/(dx)-y=0`

Answer

To find the differential equation corresponding to the family of curves given by \( y = e^x (ax + b) \), we will follow these steps:

### Step 1: Differentiate the equation once
We start with the equation:
\[ y = e^x (ax + b) \]
Now, we differentiate \( y \) with respect to \( x \) using the product rule:
\[ \frac{dy}{dx} = e^x \cdot (a) + (ax + b) \cdot e^x \]
This simplifies to:
\[ \frac{dy}{dx} = e^x (a + ax + b) = e^x (ax + a + b) \]

### Step 2: Differentiate the equation again
Next, we differentiate \( \frac{dy}{dx} \) to find the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( e^x (ax + a + b) \right) \]
Using the product rule again:
\[ \frac{d^2y}{dx^2} = e^x (ax + a + b) + e^x (a) = e^x (ax + a + b + a) = e^x (ax + 2a + b) \]

### Step 3: Express \( a \) in terms of \( y \) and \( \frac{dy}{dx} \)
From the first derivative, we can express \( a \):
\[ a = \frac{dy}{dx} \cdot e^{-x} - (ax + b) \cdot e^{-x} \]
Substituting \( y \) back into the equation gives:
\[ a = \frac{dy}{dx} e^{-x} - \left(\frac{y}{e^x} - b\right) \]
This simplifies to:
\[ a = \frac{dy}{dx} e^{-x} - \frac{y}{e^x} + b e^{-x} \]

### Step 4: Substitute \( a \) back into the second derivative
Now we substitute \( a \) back into the second derivative:
\[ \frac{d^2y}{dx^2} = e^x \left( x \left(\frac{dy}{dx} e^{-x} - \frac{y}{e^x} + b e^{-x}\right) + 2\left(\frac{dy}{dx} e^{-x} - \frac{y}{e^x} + b e^{-x}\right) + b\right) \]

### Step 5: Simplify the equation
After substituting and simplifying, we arrive at the differential equation:
\[ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0 \]

### Final Result
Thus, the required differential equation corresponding to the family of curves \( y = e^x (ax + b) \) is:
\[ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0 \]