Q10E - 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.

$\left\{\mathrm{sinx}, \mathrm{cosx}, \mathrm{tanx}\right\}$ on $\left(-\frac{\pi }{2}, \frac{\pi }{2}\right)$

Step-by-Step Solution

Answer

The function $\left\{\mathbf{sinx}\mathbf{,} \mathbf{cosx}\mathbf{,} \mathbf{tanx}\right\}$ is linearly independent on $\left(\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \frac{\mathbf{\pi }}{\mathbf{2}}\right)$.

Step 1:Using the concept of Wronskian

The given function is $\left\{\mathbf{sinx}\mathbf{,} \mathbf{cosx}\mathbf{,} \mathbf{tanx}\right\}.$

 

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,

 

$W\left[\sin x,\cos x,\tan x\right]=\left|\begin{array}{ccc}\sin x& \cos x& \tan x\\ \cos x& -\sin x& se{c}^{2}x\\ -\sin x& -\cos x& 2se{c}^{2}x\tan x\end{array}\right|$

 

Solve the above equation,

 

$$\begin{gathered} \mathbf{W}\left[\mathbf{sinx}\mathbf{,} \mathbf{cosx}\mathbf{,} \mathbf{tanx}\right]\mathbf{=}\mathbf{sinx}\left(\mathbf{-}\mathbf{2}{\mathbf{sinxsec}}^{\mathbf{2}}\mathbf{xtanx}\mathbf{+}{\mathbf{cosxsec}}^{\mathbf{2}}\mathbf{x}\right)\mathbf{-}\mathbf{cosx}\left(\mathbf{2}{\mathbf{cosxsec}}^{\mathbf{2}}\mathbf{xtanx}\mathbf{+}{\mathbf{sinxsec}}^{\mathbf{2}}\mathbf{x}\right) \\ \mathbf{+}\mathbf{tanx}\left(\mathbf{-}{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right) \\ \mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}{\mathbf{xsec}}^{\mathbf{2}}\mathbf{xtanx}\mathbf{+}{\mathbf{sinxcosxsec}}^{\mathbf{2}}\mathbf{x}\mathbf{-}\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}{\mathbf{xsec}}^{\mathbf{2}}\mathbf{xtanx} \\ \mathbf{-}{\mathbf{sinxcosxsec}}^{\mathbf{2}}\mathbf{x}\mathbf{-}\mathbf{tanx} \\ \mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{sec}}^{\mathbf{2}}\mathbf{xtanx}\mathbf{-}\mathbf{tanx} \\ \mathbf{=}\mathbf{-}\mathbf{tanx}\left(\mathbf{2}{\mathbf{sec}}^{\mathbf{2}}\mathbf{x}\mathbf{-}\mathbf{1}\right) \\ \mathbf{=}\mathbf{-}\mathbf{tanx}\left(\mathbf{2}{\mathbf{tan}}^{\mathbf{2}}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{-}\mathbf{1}\right) \\ \mathbf{=}\mathbf{-}\mathbf{tanx}\left(\mathbf{2}{\mathbf{tan}}^{\mathbf{2}}\mathbf{x}\mathbf{+}\mathbf{1}\right) \end{gathered}$$

Step 2: Check the linearly independent or dependent

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

Therefore, the function $\left\{\mathbf{sinx}\mathbf{,} \mathbf{cosx}\mathbf{,} \mathbf{tanx}\right\}$ is linearly independent on $\left(\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \frac{\mathbf{\pi }}{\mathbf{2}}\right)$.