Q13E - 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.

$\left\{x, x^2, x^3, x^4\right\}$ on $\left(-\infty , \infty \right)$

Step-by-Step Solution

Answer

Linearly independent

Step 1:Using the concept of Wronskian

The given function is $\left\{\mathbf{x}\mathbf{,} {\mathbf{x}}^{\mathbf{2}}\mathbf{,} {\mathbf{x}}^{\mathbf{3}}\mathbf{,} {\mathbf{x}}^{\mathbf{4}}\right\}.$

 

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,

 

$W\left[x,x^2,x^3,x^4\right]=\left|\begin{array}{cccc}x& x^2& x^3& x^4\\ 1& 2x& 3x^2& 4x^3\\ 0& 2& 6x& 12x^2\\ 0& 0& 6& 24x\end{array}\right|$

 

Solve the above equation,

 

$$\begin{gathered} W\left[x, x^2, x^3, x^4\right]=x\left|\begin{array}{ccc}2x& 3x^2& 4x^3\\ 1& 6x& 12x^2\\ 0& 6& 24x\end{array}\right|-x^2\left|\begin{array}{ccc}1& 3x^2& 4x^3\\ 0& 6x& 12x^2\\ 0& 6& 24x\end{array}\right|+x^3\left|\begin{array}{ccc}1& 2x& 4x^3\\ 0& 1& 12x^2\\ 0& 0& 24x\end{array}\right|-x^4\left|\begin{array}{ccc}1& 2x& 3x^2\\ 0& 1& 6x\\ 0& 0& 6\end{array}\right| \\ =x\left(96x^3\right)-x^2\left(72x^2\right)+x^3\left(24x\right)-x^4\left(6\right) \\ =96x^4-72x^4+24x^4-6x^4 \\ =42x^4 \end{gathered}$$

Step 2: Check the linearly independent or dependent

The above function is not equal to zero $\left(\forall \mathbf{x}\right)$

 

Therefore, $\left\{\mathbf{x}\mathbf{,} {\mathbf{x}}^{\mathbf{2}}\mathbf{,} {\mathbf{x}}^{\mathbf{3}}\mathbf{,} {\mathbf{x}}^{\mathbf{4}}\right\}$ are linearly independent on $\left(\mathbf{-}\infty \mathbf{,} \infty \right)$.