Question

Let $y_{1}=e^{-x} cosx, y_{2}=e^{-x} sinx$. Choose the correct statements out of the following:
A. the Wronskian determinant is zero
B. the Wronskian determinant is $e^{-2x}$}
C. $y_{1}$ and $y_{2}$ are linearly dependent}
D. $y_{1}$ and $y_{2}$ are linearly independent} Choose the correct answer from the options given below:

• A. A, C only
• B. A, D only
• C. B, C only
• D. B, D only

Hint:

For functions of the form $e^{\alpha x}\cos \beta x$ and $e^{\alpha x}\sin \beta x$, the Wronskian is always $\beta e^{2\alpha x}$. Here $\alpha = -1, \beta = 1$, so $W = 1 \cdot e^{-2x}$.

Answer 1

The Correct Option is D

Concept: The Wronskian $W(y_1, y_2)$ of two functions is used to determine linear independence. If $W \neq 0$ on an interval, the functions are linearly independent.

Step 1: Find the derivatives.
$y_1 = e^{-x} \cos x \Rightarrow y_1' = -e^{-x} \cos x - e^{-x} \sin x = -e^{-x}(\cos x + \sin x)$ $y_2 = e^{-x} \sin x \Rightarrow y_2' = -e^{-x} \sin x + e^{-x} \cos x = e^{-x}(\cos x - \sin x)$

Step 2: Calculate the Wronskian.
$W = y_1 y_2' - y_2 y_1'$ $W = (e^{-x} \cos x)[e^{-x}(\cos x - \sin x)] - (e^{-x} \sin x)[-e^{-x}(\cos x + \sin x)]$ $W = e^{-2x}(\cos^2 x - \cos x \sin x) + e^{-2x}(\sin x \cos x + \sin^2 x)$ $W = e^{-2x}(\cos^2 x - \cos x \sin x + \sin x \cos x + \sin^2 x)$ $W = e^{-2x}(\cos^2 x + \sin^2 x) = e^{-2x}(1) = e^{-2x}$.

Step 3: Analyze linear independence.
Since $W = e^{-2x}$ is never zero for any real $x$, the functions $y_1$ and $y_2$ are Linearly Independent. Conclusion: Statements B and D are correct.