Wronskian Independence 8531C7

1. **State the problem:** We want to prove that the set of functions {$e^{ax}$, $e^{bx}$} is linearly independent on $\mathbb{R}$ using the Wronskian method. 2. **Recall the Wronskian definition:** For two functions $f(x)$ and $g(x)$, the Wronskian $W(f,g)(x)$ is defined as: $$ W(f,g)(x) = \begin{vmatrix} f(x) & g(x) \\ f'(x) & g'(x) \end{vmatrix} = f(x)g'(x) - f'(x)g(x) $$ If $W(f,g)(x) \neq 0$ for some $x$ in the interval, then $f$ and $g$ are linearly independent on that interval. 3. **Apply to our functions:** Let $f(x) = e^{ax}$ and $g(x) = e^{bx}$. 4. **Compute derivatives:** $$ f'(x) = a e^{ax}, \quad g'(x) = b e^{bx} $$ 5. **Calculate the Wronskian:** $$ W(f,g)(x) = e^{ax} \cdot b e^{bx} - a e^{ax} \cdot e^{bx} = (b - a) e^{ax} e^{bx} = (b - a) e^{(a+b)x} $$ 6. **Analyze the Wronskian:** - Since $e^{(a+b)x} \neq 0$ for all $x \in \mathbb{R}$, - The Wronskian $W(f,g)(x) = 0$ if and only if $b - a = 0$, i.e., $a = b$. 7. **Conclusion:** - If $a \neq b$, then $W(f,g)(x) \neq 0$ for all $x$, so $e^{ax}$ and $e^{bx}$ are linearly independent on $\mathbb{R}$. - If $a = b$, the functions are not distinct and thus not linearly independent. **Final answer:** The set {$e^{ax}$, $e^{bx}$} is linearly independent on $\mathbb{R}$ if and only if $a \neq b$.