If `y=e^(-x) cos x` and `y_n+k_ny=0` where `yn=(d^ny)/(dx^n)` and `k_n` are constant `n in N` then

Options

A. `k_(4)=4`
B. `k_(8)=-16`
C. `k_(12)=20`
D. `k_(16)=-24`

Explanation

To solve the problem where \( y = e^{-x} \cos x \) and \( y_n + k_n y = 0 \) (where \( y_n = \frac{d^n y}{dx^n} \) and \( k_n \) are constants), we will differentiate \( y \) multiple times and find the relationship between the derivatives and \( y \).

### Step 1: Find the first derivative \( y_1 \)

Using the product rule:
\[ y_1 = \frac{d}{dx}(e^{-x} \cos x) = e^{-x} \frac{d}{dx}(\cos x) + \cos x \frac{d}{dx}(e^{-x}) \]
Calculating the derivatives:
\[ \frac{d}{dx}(\cos x) = -\sin x \quad \text{and} \quad \frac{d}{dx}(e^{-x}) = -e^{-x} \]
Substituting these into the product rule:
\[ y_1 = e^{-x}(-\sin x) + \cos x(-e^{-x}) = -e^{-x} \sin x - e^{-x} \cos x = -e^{-x}(\sin x + \cos x) \]

### Step 2: Find the second derivative \( y_2 \)

Now we differentiate \( y_1 \):
\[ y_2 = \frac{d}{dx}(-e^{-x}(\sin x + \cos x)) \]
Using the product rule again:
\[ y_2 = -\left( e^{-x} \frac{d}{dx}(\sin x + \cos x) + (\sin x + \cos x) \frac{d}{dx}(e^{-x}) \right) \]
Calculating the derivative of \( \sin x + \cos x \):
\[ \frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x \]
Substituting this back:
\[ y_2 = -\left( e^{-x}(\cos x - \sin x) - e^{-x}(\sin x + \cos x) \right) \]
Simplifying:
\[ y_2 = -e^{-x}(\cos x - \sin x) + e^{-x}(\sin x + \cos x) = -e^{-x}(-2\sin x) = 2e^{-x} \sin x \]

### Step 3: Find the third derivative \( y_3 \)

Now differentiate \( y_2 \):
\[ y_3 = \frac{d}{dx}(2e^{-x} \sin x) \]
Using the product rule:
\[ y_3 = 2\left( e^{-x} \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}(e^{-x}) \right) \]
Calculating:
\[ \frac{d}{dx}(\sin x) = \cos x \quad \text{and} \quad \frac{d}{dx}(e^{-x}) = -e^{-x} \]
Substituting:
\[ y_3 = 2\left( e^{-x} \cos x - \sin x e^{-x} \right) = 2e^{-x}(\cos x - \sin x) \]

### Step 4: Find the fourth derivative \( y_4 \)

Now differentiate \( y_3 \):
\[ y_4 = \frac{d}{dx}(2e^{-x}(\cos x - \sin x)) \]
Using the product rule:
\[ y_4 = 2\left( e^{-x} \frac{d}{dx}(\cos x - \sin x) + (\cos x - \sin x) \frac{d}{dx}(e^{-x}) \right) \]
Calculating:
\[ \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x \]
Substituting:
\[ y_4 = 2\left( e^{-x}(-\sin x - \cos x) - e^{-x}(\cos x - \sin x) \right) \]
Simplifying:
\[ y_4 = 2e^{-x}(-2\cos x) = -4e^{-x} \cos x \]

### Step 5: Relate \( y_4 \) to \( y \)

We know:
\[ y = e^{-x} \cos x \]
Thus:
\[ y_4 = -4y \]
This gives us the relationship:
\[ y_4 + 4y = 0 \]
From this, we can identify that \( k_4 = 4 \).

### Step 6: Find \( k_8 \)

Continuing this process, we can derive that:
\[ y_8 = 16y \]
Thus:
\[ y_8 - 16y = 0 \implies k_8 = -16 \]

### Final Answer
The values we found are:
- \( k_4 = 4 \)
- \( k_8 = -16 \)