Q35E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Linear Dependence of Three Functions.

 

Three functions y₁(t), y₂(t) and y₃(t) are said to be linearly dependent on an interval if, on l, at least one of these functions is a linear combination of the remaining two e.g., if y₁(t) = c₁y₂(t) + c₂y₃(t). Equivalently (compare Problem), y₁,y₂ and y₃ are linearly dependent on l if there exist constants C₁,C₂ and C₃, not all zero, such that C₁y₁(t) + C₂y₂(t) + C₃y₃(t) = 0 for all t in l. Otherwise, we say that these functions are linearly independent on. For each of the following, determine whether the given three functions are linearly dependent or linearly independent on $\left(-\infty ,\infty \right)$:

$\begin{array}{rcl}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{ }{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& \mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{t}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{.}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{ }{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& \mathbf{-}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{5}{\mathbf{sin}}^{\mathbf{2}}\mathbf{t}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{cos}}^{\mathbf{2}}\mathbf{t}\mathbf{.}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{ }{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& {\mathbf{e}}^{\mathbf{t}}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{te}}^{\mathbf{t}}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{.}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{ }{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{=}& {\mathbf{e}}^{\mathbf{t}}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}\mathbf{ }\mathbf{ }{\mathbf{y}}_{\mathbf{3}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{cosht}\mathbf{.}\end{array}$

Step-by-Step Solution

Answer

$\begin{array}{rcl}\left(\mathrm{a}\right) {\mathrm{y}}_1\left(\mathrm{t}\right)& =& 1, {\mathrm{y}}_2\left(\mathrm{t}\right)=\mathrm{t}, {\mathrm{y}}_3\left(\mathrm{t}\right)={\mathrm{t}}^2 \mathrm{are} \mathrm{linearly} \mathrm{indepandent}.\\ \left(\mathrm{b}\right) {\mathrm{y}}_1\left(\mathrm{t}\right)& =& -3, {\mathrm{y}}_2\left(\mathrm{t}\right)=5{\sin }^2\mathrm{t}, {\mathrm{y}}_3\left(\mathrm{t}\right)={\cos }^2\mathrm{t} \mathrm{are} \mathrm{inlearly} \mathrm{dependent}.\\ \left(\mathrm{c}\right) {\mathrm{y}}_1\left(\mathrm{t}\right)& =& {\mathrm{e}}^{\mathrm{t}}, {\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{te}}^{\mathrm{t}}, {\mathrm{y}}_3\left(\mathrm{t}\right)={\mathrm{t}}^2{\mathrm{e}}^{\mathrm{t}} \mathrm{are} \mathrm{inlearly} \mathrm{dependent}.\\ \left(\mathrm{d}\right) {\mathrm{y}}_1\left(\mathrm{t}\right)& =& {\mathrm{e}}^{\mathrm{t}}, {\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}, {\mathrm{y}}_3\left(\mathrm{t}\right)=\mathrm{cosht} \mathrm{are} \mathrm{inlearly} \mathrm{dependent}.\end{array}$

Step 1: Introduction

Three functions y₁,y₂ and y₃ are linearly dependent on an interval I if there are some constants c₁,c₂ such that y_i(t) = c₁y_j(t) +c₂y_k(t) for $\mathrm{t} \in \mathrm{l}$, where $i,j,k\in \{1,2,3\}$and $\mathrm{i}\ne \mathrm{j}, \mathrm{j}\ne \mathrm{k}$and $\mathrm{k}\ne \mathrm{i}$. Three functions are linearly independent if they are not linearly dependent.

Step 2: Check the equality is true for all t ∈ ( - ∞ ∞ )

Assume that those three functions are linearly dependent, i.e., there are some constants c₁,c₂ such that ${\mathit{y}}_{\mathbf{3}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}\mathbf{\iff }{\mathit{t}}^{\mathbf{2}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{t}$

 

But in the last equation, we have a polynomial of degree 2 on the left side and a polynomial of degree 1 on the right side, so there are no constants c₁,c₂ such that this equality is true for all $\in (-\infty ,\infty )$. We come to the same conclusion in cases t = c₁ + c₂t² and 1 = c₁t + c₂t². So, those functions are linearly independent.

Step 3: Finding c 1 ,c 2

From the trigonometry identity sin²t + cos²t = 1, we have that the function y(t) can be transformed to y₃(t) = 1-sin2t for all $\in (-\infty ,\infty )$.

 

Now, 

$\begin{array}{rcl}{y}_3\left(t\right)& =& 1-{\sin }^{2}t\\ & =& \frac{-3}{-3}-\frac{5}{5}{\sin }^{2}t\\ & =& \frac{1}{-3}\times (-3)-\frac{1}{5}\times \left(5{\sin }^{2}t\right)\\ & =& -\frac{1}{3}{y}_1\left(t\right)-\frac{1}{5}{y}_2\left(t\right)\end{array}$

Therefore, if we take $c_1=-\frac{1}{3}$ and $c_2=-\frac{1}{5}$ one has that ${\mathit{y}}_{\mathbf{3}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$ for $\in (-\infty ,\infty )$, so the given functions are linearly dependent.

Step 4: Check the equality is true for all t ∈ ( - ∞ , ∞ )

Assuming that those functions are linearly dependent, without loss of generality, one has that there are some constants c₁,c₂ such that ${\mathit{y}}_{\mathbf{3}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{\iff }{\mathit{t}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{t}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{t}{\mathit{e}}^{\mathbf{t}}$.

 

Dividing the previous equation by $e^t\left(e^t\ne 0\right)$ we will get t² = c₁ + c₂t, and as in part (a), one has that there are no constants c₁,c₂ such that the previous equality is true for all $\in (-\infty ,\infty )$, so our assumption that those functions are linearly dependent is wrong.

Those functions are linearly independent.

Step 5: Finding c 1 , c 2

$\cos ht=\frac{e^t+e^{-t}}{2}$ (Definition of cosht )

If one takes constant $c_1=c_2=\frac{1}{2}$ one will have that y₃(t) = c₁y₁(t) + c₂y₂(t), $\mathrm{t}\in (-\infty ,\infty )$ so those three functions are linearly dependent.