If ` y= x^(2) e^(x) ,then ( d^(2)y)/(dx^(2)) -(dy)/(dx) =`

Options

A. ` (x+1) e^(x)`
B. ` -(x+1) e^(x) `
C. ` 2(x+1) e^(x)`
D. ` -2( x+1) e^(x) `

Answer

To solve the problem where \( y = x^2 e^x \), we need to find the second derivative \( \frac{d^2y}{dx^2} \) and the first derivative \( \frac{dy}{dx} \), and then compute \( \frac{d^2y}{dx^2} - \frac{dy}{dx} \).

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Using the product rule, which states that if \( u = x^2 \) and \( v = e^x \), then:

\[ \frac{dy}{dx} = u'v + uv' \]

where \( u' = \frac{d}{dx}(x^2) = 2x \) and \( v' = \frac{d}{dx}(e^x) = e^x \).

Thus, we have:

\[ \frac{dy}{dx} = (2x)(e^x) + (x^2)(e^x) = e^x(2x + x^2) \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^x(2x + x^2)) \]

Using the product rule again, let \( u = e^x \) and \( v = 2x + x^2 \):

\[ \frac{d^2y}{dx^2} = u'v + uv' \]

Here, \( u' = e^x \) and \( v' = \frac{d}{dx}(2x + x^2) = 2 + 2x \).

So we have:

\[ \frac{d^2y}{dx^2} = e^x(2x + x^2) + e^x(2 + 2x) = e^x((2x + x^2) + (2 + 2x)) = e^x(x^2 + 4x + 2) \]

### Step 3: Compute \( \frac{d^2y}{dx^2} - \frac{dy}{dx} \)

Now, we subtract \( \frac{dy}{dx} \) from \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} - \frac{dy}{dx} = e^x(x^2 + 4x + 2) - e^x(2x + x^2) \]

Factoring out \( e^x \):

\[ = e^x \left((x^2 + 4x + 2) - (2x + x^2)\right) \]

Simplifying the expression inside the parentheses:

\[ = e^x \left(x^2 + 4x + 2 - 2x - x^2\right) = e^x(2x + 2) \]

Thus, we can factor out a 2:

\[ = 2e^x(x + 1) \]

### Final Answer:

\[ \frac{d^2y}{dx^2} - \frac{dy}{dx} = 2e^x(x + 1) \]

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