Q36E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Using the definition in Problem 35, prove that if r₁ , r₂ and r₃ are distinct real numbers, then the functions ${\mathit{e}}^{{\mathit{r}}_{\mathbf{1}}\mathit{t}}\mathbf{,}{\mathit{e}}^{{\mathit{r}}_{\mathbf{2}}\mathit{t}}$, and ${\mathit{e}}^{{\mathit{r}}_{\mathbf{3}}\mathit{t}}$ are linearly independent on $\mathbf{(}\mathbf{-}\infty \mathbf{,}\infty \mathbf{)}$.

[Hint: Assume to the contrary that, say, ${\mathit{e}}^{{\mathit{r}}_{\mathbf{1}}\mathit{t}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{{\mathit{r}}_{\mathbf{2}}\mathit{t}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{{\mathit{r}}_{\mathbf{3}}\mathit{t}}$ for all t. Divide by ${\mathit{e}}^{{\mathit{r}}_{\mathbf{2}}\mathit{t}}$ to get ${\mathit{e}}^{\left({\mathit{r}}_{\mathbf{1}}\mathbf{-}{\mathit{r}}_{\mathbf{2}}\right)\mathit{t}}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\left({\mathit{r}}_{\mathbf{3}}\mathbf{-}{\mathit{r}}_{\mathbf{2}}\right)\mathit{t}}$ and then differentiate to deduce that ${\mathit{e}}^{\left({\mathit{r}}_{\mathbf{1}}\mathbf{-}{\mathit{r}}_{\mathbf{2}}\right)\mathit{t}}$ and ${\mathit{e}}^{\left({\mathit{r}}_{\mathbf{3}}\mathbf{-}{\mathit{r}}_{\mathbf{2}}\right)\mathit{t}}$ are linearly dependent, which is a contradiction. (Why?)]

Step-by-Step Solution

Answer

Assuming that there are constants $a,b,c$ such that $a{e}^{r_{1}t}+b{e}^{r_{2}t}+c{e}^{r_{3}t}=0$ one will get that $\mathit{a}\mathbf{=}\mathit{b}\mathbf{=}\mathit{c}\mathbf{=}0$, so there are no non-zero constants $a,b,c$ such that $a{e}^{r_{1}t}+b{e}^{r_{2}t}+c{e}^{r_{3}t}=0$, therefore the given functions are linearly independent on $(-\infty ,\infty )$.

Step 1: Multiply and differentiate the equation

Let  $r_1,r_2$ and $r_3$ be distinct real numbers and consider the functions $f_1\left(t\right)=e^{r_{1}t},f_2\left(t\right)=e^{r_{2}t}$ and $f_3\left(t\right)=e^{r_{3}t}$ on $(-\infty ,\infty )$.

 

Let a, b and c be real constants such that $a{e}^{r_{1}t}+b{e}^{r_{2}t}+c{e}^{r_{3}t}=0$ for all $t\in (-\infty ,\infty )$.

 

Since $e^{-r_{1}t}\ne 0$ for all, $t\in (-\infty ,\infty )$ we can multiply both sides of $a{e}^{r_{1}t}+b{e}^{r_{2}t}+c{e}^{r_{3}t}=0$ by $e^{-r_{1}t}$ obtaining $a+b{e}^{\left(r_2-r_1\right)t}+c{e}^{\left(r_3-r_1\right)t}=0$ 

 

Differentiating both sides of $a+b{e}^{\left(r_2-r_1\right)t}+c{e}^{\left(r_3-r_1\right)t}=0$ gives that $\left(r_2-r_1\right)b{e}^{\left(r_2-r_1\right)t}+c\left(r_3-r_1\right)e^{\left(r_3-r_1\right)t}=0$ for all $t\in (-\infty ,\infty )$.

Step 2: Multiply and differentiate the equation

Since $e^{\left(r_1-r_2\right)t}\ne 0$ for all, $t\in (-\infty ,\infty )$ we can multiply both sides of

$\left(r_2-r_1\right)b{e}^{\left(r_2-r_1\right)t}+c\left(r_3-r_1\right)e^{\left(r_3-r_1\right)t}=0$ by $e^{\left(r_1-r_2\right)t}$ obtaining $\left(r_2-r_1\right)b+c\left(r_3-r_1\right)e^{\left(r_3-r_2\right)t}=0$.

 

Differentiating both sides of $\left(r_2-r_1\right)b+c\left(r_3-r_1\right)e^{\left(r_3-r_2\right)t}=0$ gives that $c\left(r_3-r_1\right)\left(r_3-r_2\right)e^{\left(r_3-r_2\right)t}=0$

 for all $t\in (-\infty ,\infty )$.

 

Hence, $\left(r_3-r_1\right)\left(r_3-r_2\right)\ne 0$ and ${\mathit{e}}^{\left({\mathit{r}}_{\mathbf{3}}\mathbf{-}{\mathit{r}}_{\mathbf{2}}\right)\mathit{t}}\ne 0$ for all $t\in (-\infty ,\infty )$ one has that $c=0$

Step 3: Finding the general solution

Then, $\left(r_2-r_1\right)b{e}^{\left(r_2-r_1\right)t}+c\left(r_3-r_1\right)e^{\left(r_3-r_1\right)t}=0$ for all $t\in (-\infty ,\infty )$ one has that $\left(r_2-r_1\right)b{e}^{\left(r_2-r_1\right)t}=0$ for all $t\in (-\infty ,\infty )$ and therefore since $\left(r_2-r_1\right)\ne 0$ and $e^{\left(r_2-r_1\right)t}\ne 0$ for all $t\in (-\infty ,\infty )$ one has that $b=0$.

 

Thus, $a+b{e}^{\left(r_2-r_1\right)t}+c{e}^{\left(r_3-r_1\right)t}=0$ for all, $t\in (-\infty ,\infty )$ we have that $a=0$.

 

Hence, $a=b=c=0$ and therefore there are no constants a, b and c not all 0 such that $a{f}_1\left(t\right)+b{f}_2\left(t\right)+c{f}_3\left(t\right)=0$ for all $t\in (-\infty ,\infty )$ then $f_1\left(t\right),f_2\left(t\right)$ and $f_3\left(t\right)$ are linearly independent on $(-\infty ,\infty )$.