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Q26E

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Found in: Page 332

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# As an alternative proof that the functions ${e}^{{r}_{1}x},{e}^{{r}_{2}x},{e}^{{r}_{3}x},...,{e}^{{r}_{n}x}$ are linearly independent on (∞,-∞) when ${{r}}_{{1}}{,}{{r}}_{{2}}{,}{.}{..}{{r}}_{{n}}$ are distinct, assume ${{C}}_{{1}}{{e}}^{{r}_{1}x}{+}{{C}}_{{2}}{{e}}^{{r}_{2}x}{+}{{C}}_{{3}}{{e}}^{{r}_{3}x}{+}{.}{..}{+}{{C}}_{{n}}{{e}}^{{r}_{n}x}$ holds for all x in (∞,-∞) and proceed as follows:(a) Because the ri’s are distinct we can (if necessary)relabel them so that ${{r}}_{{1}}{>}{{r}}_{{2}}{>}{{r}}_{{3}}{>}{.}{..}{>}{{r}}_{{n}}$.Divide equation (33) by to obtain ${{C}}_{{1}}{+}\frac{{C}_{2}{e}^{{r}_{2}x}}{{e}^{{r}_{2}x}}{+}\frac{{C}_{3}{e}^{{r}_{3}x}}{{e}^{{r}_{3}x}}{+}{.}{..}{+}\frac{{C}_{n}{e}^{{r}_{n}x}}{{e}^{{r}_{n}x}}{=}{0}$ Now let x→∞ on the left-hand side to obtain C1 = 0. (b) Since C1 = 0, equation (33) becomes ${{C}}_{{2}}{{e}}^{{r}_{2}x}{+}{{C}}_{{3}}{{e}}^{{r}_{3}x}{+}{.}{..}{+}{{C}}_{{n}}{{e}}^{{r}_{n}x}$ = 0for all x in(∞,-∞). Divide this equation by ${{e}}^{{r}_{2}x}$and let x→∞ to conclude that C2 = 0.(c) Continuing in the manner of (b), argue that all thecoefficients, C1, C2, . . . ,Cn are zero and hence ${{e}}^{{r}_{1}x}{,}{{e}}^{{r}_{2}x}{,}{{e}}^{{r}_{3}x}{,}{.}{..}{,}{{e}}^{{r}_{n}x}$are linearly independent on (∞,-∞).

The answer to this problem is:

${C}_{1}=0$, ${C}_{2}=0$

See the step by step solution

## Step 1: Basic differentiation

The Sum rule says the derivative of a sum of functions is the sum of their derivatives. The Difference rule says the derivative of a difference of functions is the difference of their derivatives.

## Step 2: Solving by basic differentiation:

We will do the following question on the basis of basic differentiation ;

$\begin{array}{l}\left\{{e}^{{r}_{1}x},{e}^{{r}_{2}x},{e}^{{r}_{3}x},...,{e}^{{r}_{n}x}\right\}\\ {r}_{i}\ne {r}_{j}\forall i,j\notin \left\{1,2,3,...,n\right\}\\ \\ {C}_{1}{e}^{{r}_{1}x}+{C}_{2}{e}^{{r}_{2}x}+{C}_{3}{e}^{{r}_{3}x}+...+{C}_{n}{e}^{{r}_{n}x}\end{array}$

(a)

$\begin{array}{l}{C}_{1}{e}^{{r}_{1}x}+{C}_{2}{e}^{{r}_{2}x}+{C}_{3}{e}^{{r}_{3}x}+...+{C}_{n}{e}^{{r}_{n}x}\\ {r}_{1}>{r}_{2}>{r}_{3}>...>{r}_{n}\\ \frac{{C}_{1}{e}^{{r}_{1}x}}{{e}^{{r}_{1}x}}+\frac{{C}_{2}{e}^{{r}_{2}x}}{{e}^{{r}_{2}x}}+\frac{{C}_{3}{e}^{{r}_{3}x}}{{e}^{{r}_{3}x}}+...+\frac{{C}_{n}{e}^{{r}_{n}x}}{{e}^{{r}_{n}x}}=0\\ {C}_{1}+{C}_{2}\left({e}^{\left({r}_{2}-{r}_{1}\right)x}\right)+{C}_{3}\left({e}^{\left({r}_{3}-{r}_{1}\right)x}\right)+...+{C}_{n}\left({e}^{\left({r}_{n}-{r}_{1}\right)x}\right)=0\\ \\ {e}^{\left({r}_{2}-{r}_{1}\right)x}={e}^{-{m}_{2}x}\\ {e}^{\left({r}_{3}-{r}_{1}\right)x}={e}^{-{m}_{3}x}\\ {e}^{\left({r}_{n}-{r}_{1}\right)x}={e}^{-{m}_{n}x}\\ \\ \underset{x\to \infty }{\mathrm{lim}}\left({C}_{1}+\frac{{C}_{2}}{{e}^{{m}_{2}x}}+\frac{{C}_{3}}{{e}^{{m}_{3}x}}+....+\frac{{C}_{n}}{{e}^{{m}_{n}x}}\right)=0\\ {e}^{x}\to \infty \\ x\to +\infty \\ {C}_{1}+{C}_{2}\left(0\right)+....+{C}_{n}\left(0\right)=0\\ {C}_{1}=0\end{array}$

(b)

$\begin{array}{l}{C}_{2}{e}^{{r}_{2}x}+{C}_{3}{e}^{{r}_{3}x}+...+{C}_{n}{e}^{{r}_{n}x}\\ {C}_{2}+{C}_{3}\frac{{e}^{{r}_{3}x}}{{e}^{{r}_{2}x}}+...+{C}_{n}\frac{{e}^{{r}_{n}x}}{{e}^{{r}_{3}x}}=0\\ {C}_{2}+{C}_{3}\left({e}^{\left({r}_{3}-{r}_{1}\right)x}\right)+...+{C}_{n}\left({e}^{\left({r}_{n}-{r}_{1}\right)x}\right)=0\\ {e}^{\left({r}_{3}-{r}_{1}\right)x}={e}^{-{s}_{3}}\\ {e}^{\left({r}_{n}-{r}_{1}\right)x}={e}^{-{s}_{n}}\\ \\ \underset{x\to \infty }{\mathrm{lim}}\left({C}_{2}+{C}_{3}\left({e}^{-{s}_{3}x}\right)+{C}_{4}\left({e}^{-{s}_{4}x}\right)+...+{C}_{n}\left({e}^{-{s}_{n}x}\right)\right)=0\\ {e}^{x}\to \infty \\ x\to +\infty \\ {C}_{2}+{C}_{3}\left(0\right)+....+{C}_{n}\left(0\right)=0\\ {C}_{2}=0\\ \end{array}$

(c)

${C}_{1}={C}_{2}={C}_{3}=...={C}_{n}=0$

${C}_{1}=0$ , ${C}_{2}=0$