The differential equation for `y=e^(x)(a+bx)` 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 for \( y = e^{x}(a + bx) \), we will follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = e^{x}(a + bx) \]
We will differentiate \( y \) with respect to \( x \) using the product rule. The product rule states that if you have two functions \( u \) and \( v \), then:
\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Here, let \( u = e^{x} \) and \( v = a + bx \).

### Step 2: Apply the product rule
Differentiating \( u \) and \( v \):
\[ \frac{du}{dx} = e^{x}, \quad \frac{dv}{dx} = b \]
Now applying the product rule:
\[ \frac{dy}{dx} = e^{x}(b) + (a + bx)(e^{x}) = e^{x}b + e^{x}(a + bx) \]
This simplifies to:
\[ \frac{dy}{dx} = e^{x}(b + a + bx) \]
We can rearrange this as:
\[ \frac{dy}{dx} = e^{x}(b + y) \]

### Step 3: Differentiate again
Now we differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^{x}(b + y)) \]
Using the product rule again:
\[ \frac{d^2y}{dx^2} = e^{x}(b + y) + e^{x}\frac{dy}{dx} \]
Substituting \( \frac{dy}{dx} \) from our previous step:
\[ \frac{d^2y}{dx^2} = e^{x}(b + y) + e^{x}(b + y) = 2e^{x}(b + y) \]

### Step 4: Substitute back to eliminate \( b \)
From the first derivative, we know:
\[ b = \frac{dy}{dx} - y \]
Substituting this back into our second derivative:
\[ \frac{d^2y}{dx^2} = 2e^{x}\left(\frac{dy}{dx} - y + y\right) = 2e^{x}\frac{dy}{dx} \]

### Step 5: Form the differential equation
Now we can rearrange our equation:
\[ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \]
This is our final differential equation.

### Final Result
The differential equation for \( y = e^{x}(a + bx) \) is:
\[ \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \]