Q12E - 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.

$\left\{\cos 2x, {\cos }^{2}x, {\sin }^{2}x\right\}$ on $\left(-\infty , \infty \right)$

Step-by-Step Solution

Answer

Therefore, the function $\left\{\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{,} {\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{,} {\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right\}$ is linearly dependent on $\left(\mathbf{-}\infty \mathbf{,} \infty \right)$.

Step 1: Using the concept of Wronskian

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

 

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,

 

$$\begin{gathered} W\left[\cos 2x,{\cos }^{2}x,{\sin }^{2}x\right]=W\left[\cos 2x,\frac{1+\cos 2x}{2},\frac{1-\cos 2x}{2}\right] \\ =\left|\begin{array}{ccc}\cos 2x& \frac{1+\cos 2x}{2}& \frac{1-\cos 2x}{2}\\ -2\sin 2x& -\sin 2x& \sin 2x\\ -4\cos 2x& -2\cos 2x& 2\cos 2x\end{array}\right| \end{gathered}$$

 

Solve the above equation,

$$\begin{gathered} \mathbf{W}\left[\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{,} {\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{,} {\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right]\mathbf{=}\mathbf{cos}\mathbf{2}\mathbf{x}\left(\mathbf{-}\mathbf{2}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\right) \\ \mathbf{-}\frac{\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{-}\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\right) \\ \mathbf{+}\frac{\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\right) \\ \mathbf{=}\mathbf{cos}\mathbf{2}\mathbf{x}\left(\mathbf{0}\right)\mathbf{-}\frac{\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{0}\right)\mathbf{+}\frac{\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{0}\right) \\ \mathbf{=}\mathbf{0} \end{gathered}$$

Step 2: Check the linearly independent or dependent

The above function is equal to zero $\left(\forall \mathbf{x}\right).$

 

Therefore, $\left\{\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{,} {\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{,} {\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right\}$ is linearly dependent on $\left(\mathbf{-}\infty \mathbf{,} \infty \right)$.