Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

$\left\{e^{3x}, e^{5x}, e^{-x}\right\}$ on $\left(-\infty , \infty \right)$

Step-by-Step Solution

Answer

Thus, $\left\{{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\mathbf{,} {\mathbf{e}}^{\mathbf{5}\mathbf{x}}\mathbf{,} {\mathbf{e}}^{\mathbf{-}\mathbf{x}}\right\}$ are linearly independent on $\left(\mathbf{-}\infty \mathbf{,} \infty \right)$.

Step 1:Using the concept of Wronskian

The given function is $\left\{{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\mathbf{,} {\mathbf{e}}^{\mathbf{5}\mathbf{x}}\mathbf{,} {\mathbf{e}}^{\mathbf{-}\mathbf{x}}\right\}.$

 

Apply the concept of Wronskian,

 

$W\left[f_1,f_2,\dots ,f_n\right]=\left|\begin{array}{cccc}{f}_1\left(x\right)& f_2\left(x\right)& \dots & f_n\left(x\right)\\ f_1\text{'}\left(x\right)& f_2\text{'}\left(x\right)& \dots & f_n\text{'}\left(x\right)\\ ⋮& ⋮& & ⋮\\ {f_1}^{n-1}\left(x\right)& {f_2}^{n-1}\left(x\right)& \cdots & {f_n}^{n-1}\left(x\right)\end{array}\right|$

Therefore,

 

$W\left[e^{3x},e^{5x},e^{-x}\right]=\left|\begin{array}{ccc}{e}^{3x}& e^{5x}& e^{-x}\\ 3e^{3x}& 5e^{5x}& -e^{-x}\\ 9e^{3x}& 25e^{5x}& e^{-x}\end{array}\right|$

 

Solve the above equation,

 

$$\begin{gathered} W\left[e^{3x},e^{5x},e^{-x}\right]=e^{3x}\times e^{5x}\times e^{-x}\left|\begin{array}{ccc}1& 1& 1\\ 3& 5& -1\\ 9& 25& 1\end{array}\right| \\ =e^{7x}\left[1\left(30\right)-1\left(12\right)+1\left(30\right)\right] \\ =e^{7x}\left[\left[48\right]\right] \\ =48e^{7x} \end{gathered}$$

Step 2: Check the linearly independent or dependent

Since the above result is $\mathbf{48}{\mathbf{e}}^{\mathbf{7}\mathbf{x}}\ne \mathbf{0}$ $\left(\forall \mathbf{x}\right)$.

 

Therefore, $\left\{{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\mathbf{,} {\mathbf{e}}^{\mathbf{5}\mathbf{x}}\mathbf{,} {\mathbf{e}}^{\mathbf{-}\mathbf{x}}\right\}$ are linearly independent on $\left(\mathbf{-}\infty \mathbf{,} \infty \right)$.