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 allt. 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?)]

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 $t\in (-\infty ,\infty )$.

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$c=0$

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 $(-\infty ,\infty )$.