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Q30E.

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Found in: Page 333

### 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

# (a) Derive the form ${y}{\left(}{x}{\right)}{=}{{A}}_{{1}}{{e}}^{{x}}{+}{{A}}_{{2}}{{e}}^{-x}{+}{{A}}_{{3}}{c}{o}{s}{x}{+}{{A}}_{{4}}{s}{i}{n}{x}$ for the general solution to the equation ${{y}}^{\left(4\right)}{=}{y}$, from the observation that the fourth roots of unity are 1, -1, i, and -i. (b) Derive the form ${y}{\left(}{x}{\right)}{=}{{A}}_{{1}}{{e}}^{{x}}{+}{{A}}_{{2}}{{e}}^{-x/2}{c}{o}{s}\left(\frac{\sqrt{3x}}{2}\right){+}{{A}}_{{3}}{{e}}^{-x/2}{s}{i}{n}\left(\frac{\sqrt{3x}}{2}\right)$for the general solution to the equation ${{y}}^{\left(3\right)}{=}{y}$ from the observation that the cube roots of unity are 1, ${{e}}^{i\frac{2\pi }{3}}$ , and ${{e}}^{-i\frac{2\pi }{3}}$.

Proved.

See the step by step solution

## Step 1: Solving for (a):

(a)Let differential equation be,

${y}^{\left(4\right)}=y$

Then its corresponding auxillary equation is,

$\begin{array}{l}{m}^{4}=1\\ m=1,-1,i,-i\end{array}$

Let,

$\begin{array}{l}{m}_{1}=1\\ {m}_{2}=-1\\ {m}_{3}=i\\ {m}_{4}=-i\end{array}$

Here, two roots are real and distinct and other two roots are purely complex and distinct.

Then the general solution is given by,

$y={c}_{1}{e}^{{m}_{1}x}+{c}_{2}{e}^{{m}_{2}x}+{c}_{3}{e}^{{m}_{3}x}+{c}_{4}{e}^{{m}_{4}x}$ ( ${c}_{1},{c}_{2},{c}_{3},{c}_{4}$ are constant coefficient)

$\begin{array}{l}={c}_{1}{e}^{x}+{c}_{2}{e}^{-x}+{c}_{3}{e}^{ix}+{c}_{4}{e}^{-ix}\\ ={c}_{1}{e}^{x}+{c}_{2}{e}^{-x}+{c}_{3}\left(cos\left(x\right)+isin\left(x\right)\right)+{c}_{4}\left(cos\left(-x\right)+isin\left(-x\right)\right)\\ ={c}_{1}{e}^{x}+{c}_{2}{e}^{-x}+cos\left(x\right)\left({c}_{3}+{c}_{4}\right)+sin\left(x\right)\left({c}_{3}i+{c}_{4}i\right)\end{array}$

Let ${A}_{3}={c}_{3}+{c}_{4}$ and ${A}_{4}={c}_{3}i+{c}_{4}i$

Then,

$y={c}_{1}{e}^{x}+{c}_{2}{e}^{-x}+{A}_{3}\mathrm{cos}\left(x\right)+{A}_{4}\mathrm{sin}\left(x\right)$

## Step 2: proving further:

Let ${c}_{1}={A}_{1}$ and ${c}_{2}={A}_{2}$

$y={A}_{1}{e}^{x}+{A}_{2}{e}^{-x}+{A}_{3}\mathrm{cos}\left(x\right)+{A}_{4}\mathrm{sin}\left(x\right)$

Hence proved

## Step 3: Solving for (b):

(b)Let the given differential equation be,

${y}^{\left(3\right)}=y$

Then its corresponding auxillary equation is,

$\begin{array}{l}{m}^{3}=1\\ y\text{'}=1,{e}^{\frac{2\pi i}{3}},{e}^{-\frac{2\pi i}{3}}\end{array}$

Let,

$\begin{array}{l}{m}_{1}=1\\ {m}_{2}={e}^{\frac{2\pi i}{3}}\\ {m}_{3}={e}^{-\frac{2\pi i}{3}}\end{array}$

Here, two roots are real and distinct and other two roots are purely complex and distinct.

Therefore, the general solution is given by

$y={A}_{1}{e}^{{m}_{1}x}+C{e}^{{m}_{2}x}+D{e}^{{m}_{3}x}$

Here now to find exact value of ${e}^{\frac{2\pi i}{3}}$ and ${e}^{-\frac{2\pi i}{3}}$

$\begin{array}{l}{e}^{\frac{2\pi i}{3}}=\mathrm{cos}\left(\frac{2\pi }{3}\right)+i\mathrm{sin}\left(\frac{2\pi }{3}\right)\\ =\mathrm{cos}\left(\pi -\frac{\pi }{3}\right)+i\mathrm{sin}\left(\pi -\frac{\pi }{3}\right)\\ =\mathrm{cos}\left(\pi \right)\mathrm{cos}\left(\frac{\pi }{3}\right)+\mathrm{sin}\left(\pi \right)\mathrm{sin}\left(\frac{\pi }{3}\right)+i\left(\mathrm{sin}\left(\pi \right)\mathrm{cos}\left(\frac{\pi }{3}\right)-\mathrm{cos}\left(\pi \right)\mathrm{sin}\left(\frac{\pi }{3}\right)\right)\\ =\left(-1.\frac{1}{2}+0\right)+i\left(0-\frac{\sqrt{3}}{2}.-1\right)\\ {e}^{\frac{2\pi i}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\end{array}$

On similar lines

${e}^{\frac{2\pi i}{3}}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$ …(3)

## Step 3: Solving by substituting further:

Substituting (2) and (3) in (1) we get,

$\begin{array}{l}y={A}_{1}{e}^{x}+C{e}^{\left(-\frac{1}{2}+i\frac{\sqrt{3}x}{2}\right)}+D{e}^{\left(-\frac{1}{2}-i\frac{\sqrt{3}x}{2}\right)}\\ ={A}_{1}{e}^{x}+C\left[{e}^{-\frac{1}{2}x}.{e}^{i\frac{\sqrt{3}x}{2}}\right]+D\left[{e}^{-\frac{1}{2}x}.{e}^{-i\frac{\sqrt{3}x}{2}}\right]\\ ={A}_{1}{e}^{x}+C{e}^{-\frac{1}{2}x}\left(\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right)+i\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right)\right)+D{e}^{-\frac{1}{2}x}\left(\mathrm{cos}\left(-\frac{\sqrt{3}x}{2}\right)+i\mathrm{sin}\left(-\frac{\sqrt{3}x}{2}\right)\right)\\ ={A}_{1}{e}^{x}+C{e}^{-\frac{1}{2}x}\left(\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right)+i\mathrm{sin}\left(\frac{\sqrt{3x}}{2}\right)\right)+D{e}^{-\frac{1}{2}x}\left(\mathrm{cos}\left(\frac{\sqrt{3x}}{2}\right)-i\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right)\right)\\ ={A}_{1}{e}^{x}+{e}^{-\frac{1}{2}x}\left[C\left(\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right)+i\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right)\right)+D\left(\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right)-i\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right)\right)\right]\\ ={A}_{1}{e}^{x}+{e}^{-\frac{1}{2}x}\left[\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right)\left(Ci+Di\right)+\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right)\left(Ci-Di\right)\right]\end{array}$

Let ${A}_{2}=C+D$ and ${A}_{3}=Ci-Di$

$\begin{array}{l}⇒y={A}_{1}{e}^{x}+{e}^{-\frac{1}{2}x}\left[\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right){A}_{2}+\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right){A}_{3}\right]\\ ={A}_{1}{e}^{x}+{A}_{2}{e}^{-\frac{1}{2}x}\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right)+{A}_{3}{e}^{-\frac{1}{2}x}\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right)\end{array}$

$y={A}_{1}{e}^{x}+{A}_{2}{e}^{-\frac{1}{2}x}\mathrm{cos}\left(\frac{\sqrt{3}x}{2}\right)+{A}_{3}{e}^{-\frac{1}{2}x}\mathrm{sin}\left(\frac{\sqrt{3}x}{2}\right)$

Hence proved.