If `y = e^(-x)(Acos X + Bsin x)`, then y is a solution of

Options

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

Answer

To determine the differential equation for the function \( y = e^{-x}(A \cos x + B \sin x) \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = e^{-x}(A \cos x + B \sin x) \]

Using the product rule for differentiation:
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{-x}) \cdot (A \cos x + B \sin x) + e^{-x} \cdot \frac{d}{dx}(A \cos x + B \sin x) \]

Calculating the derivatives:
- The derivative of \( e^{-x} \) is \( -e^{-x} \).
- The derivative of \( A \cos x + B \sin x \) is \( -A \sin x + B \cos x \).

So we have:
\[ \frac{dy}{dx} = -e^{-x}(A \cos x + B \sin x) + e^{-x}(-A \sin x + B \cos x) \]

This simplifies to:
\[ \frac{dy}{dx} = -e^{-x}(A \cos x + B \sin x) + e^{-x}(-A \sin x + B \cos x) \]
\[ = e^{-x}(-A \sin x + B \cos x - A \cos x - B \sin x) \]
\[ = e^{-x}(-A \sin x + B \cos x - A \cos x - B \sin x) \]

### Step 2: Substitute \( y \) into the expression

We know:
\[ y = e^{-x}(A \cos x + B \sin x) \]

Thus, we can rewrite \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -y + e^{-x}(-A \sin x + B \cos x) \]

### Step 3: Differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-y + e^{-x}(-A \sin x + B \cos x)\right) \]

Using the product rule again:
\[ \frac{d^2y}{dx^2} = -\frac{dy}{dx} + \frac{d}{dx}(e^{-x})(-A \sin x + B \cos x) + e^{-x} \frac{d}{dx}(-A \sin x + B \cos x) \]

Calculating the derivatives:
- The derivative of \( e^{-x} \) is \( -e^{-x} \).
- The derivative of \( -A \sin x + B \cos x \) is \( -A \cos x - B \sin x \).

So we have:
\[ \frac{d^2y}{dx^2} = -\frac{dy}{dx} + e^{-x}(-A \cos x - B \sin x) + e^{-x}(-A \cos x - B \sin x) \]
\[ = -\frac{dy}{dx} - 2e^{-x}(A \cos x + B \sin x) \]

### Step 4: Substitute \( \frac{dy}{dx} \) into the equation

Now we can substitute \( \frac{dy}{dx} \) back into the equation:
\[ \frac{d^2y}{dx^2} = -(-y + e^{-x}(-A \sin x + B \cos x)) - 2e^{-x}(A \cos x + B \sin x) \]
\[ = y - e^{-x}(-A \sin x + B \cos x) - 2e^{-x}(A \cos x + B \sin x) \]

### Final Differential Equation

Thus, we arrive at the final form of the differential equation:
\[ \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0 \]