Determining Linear Independence of Functions

The question asks about the nature of the solutions \(e^x\), \(e^{-x}\), and \(e^{2x}\) for the given third-order linear homogeneous differential equation: \(\frac{{{d^3}y}}{{d{x^3}}} - 2\frac{{{d^2}y}}{{d{x^2}}} - \frac{{dy}}{{dx}} + 2y = 0\). Specifically, we need to determine if these functions are linearly dependent or linearly independent on various real intervals.

Understanding Linear Independence

For a set of functions, linear independence is a fundamental concept in linear algebra and differential equations. A set of functions \(f_1(x), f_2(x), \ldots, f_n(x)\) is said to be linearly independent on an interval \(I\) if the only way to satisfy the equation:

\[ c_1 f_1(x) + c_2 f_2(x) + \ldots + c_n f_n(x) = 0 \]

for all \(x\) in \(I\) is for all the constants \(c_1, c_2, \ldots, c_n\) to be zero. If there exist constants, not all zero, such that the equation holds, then the functions are linearly dependent.

The Wronskian Method for Functions

For functions that are sufficiently differentiable (like exponential functions), a powerful tool to check for linear independence is the Wronskian determinant. For \(n\) functions \(f_1(x), f_2(x), \ldots, f_n(x)\), the Wronskian \(W(f_1, f_2, \ldots, f_n)(x)\) is defined as:

Theorem: If \(W(f_1, f_2, \ldots, f_n)(x) \neq 0\) for at least one point in an interval \(I\), then the functions \(f_1, f_2, \ldots, f_n\) are linearly independent on \(I\). Conversely, if the functions are linearly dependent, their Wronskian is zero for all \(x\) in \(I\).

Calculating the Wronskian for \(e^x, e^{-x}, e^{2x}\)

Let the given functions be:

• \(f_1(x) = e^x\)
• \(f_2(x) = e^{-x}\)
• \(f_3(x) = e^{2x}\)

We need to find their first and second derivatives:

• \(f_1'(x) = e^x\)
• \(f_1''(x) = e^x\)
• \(f_2'(x) = -e^{-x}\)
• \(f_2''(x) = e^{-x}\)
• \(f_3'(x) = 2e^{2x}\)
• \(f_3''(x) = 4e^{2x}\)

Now, we set up the Wronskian determinant:

\[ W(x) = \begin{vmatrix} e^x & e^{-x} & e^{2x} \\ e^x & -e^{-x} & 2e^{2x} \\ e^x & e^{-x} & 4e^{2x} \end{vmatrix} \]

We can factor out \(e^x\) from the first column, \(e^{-x}\) from the second column, and \(e^{2x}\) from the third column:

\[ W(x) = (e^x)(e^{-x})(e^{2x}) \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 1 & 1 & 4 \end{vmatrix} \]

Since \((e^x)(e^{-x})(e^{2x}) = e^{x-x+2x} = e^{2x}\), the expression becomes:

\[ W(x) = e^{2x} \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 1 & 1 & 4 \end{vmatrix} \]

Now, let's evaluate the 3x3 determinant:

\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 1 & 1 & 4 \end{vmatrix} = 1((-1)(4) - (2)(1)) - 1((1)(4) - (2)(1)) + 1((1)(1) - (-1)(1)) \]

\[ = 1(-4 - 2) - 1(4 - 2) + 1(1 + 1) \]

\[ = 1(-6) - 1(2) + 1(2) \]

\[ = -6 - 2 + 2 \]

\[ = -6 \]

So, the Wronskian is:

\[ W(x) = e^{2x}(-6) = -6e^{2x} \]

Conclusion on Linear Independence

The Wronskian \(W(x) = -6e^{2x}\). We know that \(e^{2x}\) is always positive for any real value of \(x\). Since \(-6\) is a non-zero constant, the product \(-6e^{2x}\) will never be zero for any real \(x\). That is, \(W(x) \neq 0\) for all \(x \in (-\infty, \infty)\).

According to the Wronskian theorem, because the Wronskian is non-zero for all real \(x\), the functions \(e^x\), \(e^{-x}\), and \(e^{2x}\) are linearly independent on every real interval.

This directly aligns with the option stating that the functions are "linearly independent on every real interval."