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Expert-verified 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 # Higher-Order Cauchy–Euler Equations. A differential equation that can be expressed in the form ${{a}}_{{n}}{{x}}^{{n}}{{y}}^{\left(n\right)}{\left(}{x}{\right)}{+}{{a}}_{n-1}{{x}}^{n-1}{{y}}^{\left(n-1\right)}{\left(}{x}{\right)}{+}{.}{..}{+}{{a}}_{{0}}{y}{\left(}{x}{\right)}{=}{0}$where ${{a}}_{{n}}{,}{{a}}_{n-1}{,}{.}{...}{,}{{a}}_{{0}}$ are constants, is called a homogeneous Cauchy–Euler equation. (The second-order case is discussed in Section 4.7.) Use the substitution ${y}{=}{{x}}^{{y}}$ to help determine a fundamental solution set for the following Cauchy–Euler equations:(a) ${{x}}^{{3}}{y}{\text{'}}{\text{'}}{\text{'}}{+}{{x}}^{{2}}{y}{\text{'}}{\text{'}}{-}{2}{x}{y}{\text{'}}{+}{2}{y}{=}{0}{,}{x}{>}{0}$(b) ${{x}}^{{4}}{{y}}^{\left(4\right)}{+}{6}{{x}}^{{3}}{y}{\text{'}}{\text{'}}{\text{'}}{+}{2}{{x}}^{{2}}{y}{\text{'}}{\text{'}}{-}{4}{x}{y}{+}{4}{y}{=}{0}{,}{x}{>}{0}$(c) ${{x}}^{{3}}{y}{\text{'}}{\text{'}}{\text{'}}{-}{2}{{x}}^{{2}}{y}{\text{'}}{\text{'}}{+}{13}{x}{y}{\text{'}}{-}{13}{y}{=}{0}{,}{x}{>}{0}$[Hint: ${{x}}^{\alpha +\beta i}{=}{{e}}^{\left(\alpha +\beta i\right)lnx}{=}{{x}}^{{a}}{\left\{}{c}{o}{s}{\left(}{\beta }{l}{n}{x}{\right)}{+}{i}{s}{i}{n}{\left(}{\beta }{l}{n}{x}{\right)}{\right\}}$]

$\left\{{x}^{1},{x}^{2}\left\{cos\left(3lnx\right)+isin\left(3lnx\right)\right\},{x}^{2}\left\{cos\left(3lnx\right)-isin\left(3lnx\right)\right\}$ is the fundamental solution set.

See the step by step solution

## Step 1: Solving for (a):

(a)Given differential equation is,

${x}^{3}y\text{'}\text{'}\text{'}+{x}^{2}y\text{'}\text{'}-2xy\text{'}+2y=0$ …(1)

Let,

$\begin{array}{l}y={x}^{r}\\ y\text{'}=r{x}^{r-1}\\ y\text{'}\text{'}=r\left(r-1\right){x}^{r-2}\\ y\text{'}\text{'}\text{'}=r\left(r-1\right)\left(r-2\right){x}^{r-3}\end{array}$

Substitution in equation (1) we get,

$\begin{array}{l}\left(r{x}^{r-1}\right)+2{x}^{r}=0\\ r\left(r-1\right)\left(r-2\right){x}^{r-3+3}+r\left(r-1\right){x}^{r-2+2}-2r{x}^{r-1+1}+2{x}^{r}=0\\ r\left(r-1\right)\left(r-2\right){x}^{r}+r\left(r-1\right){x}^{r}-2r{x}^{r}+2{x}^{r}=0\\ \left(r\left(r-1\right)\left(r-2\right)+r\left(r-1\right)-2r+2\right){x}^{r}=0\end{array}$

Since $x>0,{x}^{r}>0$

$\begin{array}{l}\left(r\left(r-1\right)\left(r-2\right)+r\left(r-1\right)-2r+2\right)=0\\ r\left(r-1\right)\left(r-2\right)+r\left(r-1\right)-2\left(r-1\right)=0\\ \left(r-1\right)\left(r\left(r-2\right)+r-2\right)=0\\ \left(r-1\right)\left(r\left(r-2\right)+\left(r-2\right)\right)=0\\ \left(r-1\right)\left(r-2\right)\left(r+1\right)=0\\ ⇒r=1,r=2,r=-1\end{array}$

Therefore,

${y}_{1}={x}^{1},{y}_{2}={x}^{2},{y}_{3}={x}^{-1}$

Are all solutions of equation (1)

$\left\{{x}^{1},{x}^{2},{x}^{-1}\right\}$ is the fundamental solution set.

## Step 2: Solution for (b):

(b)Given differential equation is,

${x}^{4}{y}^{4}+6{x}^{3}y\text{'}\text{'}\text{'}+2{x}^{2}y\text{'}\text{'}-4xy\text{'}+4y=0$ …(2)

Let,

$\begin{array}{l}y={x}^{r}\\ y\text{'}=r{x}^{r-1}\\ y\text{'}\text{'}=r\left(r-1\right){x}^{r-2}\\ y\text{'}\text{'}\text{'}=r\left(r-1\right)\left(r-2\right){x}^{r-3}\\ y\text{'}\text{'}\text{'}\text{'}=r\left(r-1\right)\left(r-2\right)\left(r-3\right){x}^{r-4}\end{array}$

Substituting in equation (2), we get

$\begin{array}{l}{x}^{4}\left(r\left(r-1\right)\left(r-2\right)\left(r-3\right){x}^{r-4}\right)+6{x}^{3}\left(r\left(r-1\right)\left(r-2\right){x}^{r-3}\right)+2{x}^{2}\left(r\left(r-1\right){x}^{r-2}\right)-4x\left(r{x}^{r-1}\right)+4\left({x}^{r}\right)=0\\ r\left(r-1\right)\left(r-2\right)\left(r-3\right){x}^{r-4+4}+6r\left(r-1\right)\left(r-3\right){x}^{r-3+3}+2r\left(r-1\right){x}^{r-2+2}-4r{x}^{r-1+1}+4{x}^{r}=0\\ r\left(r-1\right)\left(r-2\right)\left(r-3\right){x}^{r}+6r\left(r-1\right)\left(r-3\right){x}^{r}+2r\left(r-1\right){x}^{r}-4r{x}^{r}+4{x}^{r}=0\\ \left(r\left(r-1\right)\left(r-2\right)\left(r-3\right)+6r\left(r-1\right)\left(r-3\right)+2r\left(r-1\right)-4r+4\right){x}^{r}=0\end{array}$

Since $x>0,{x}^{r}>0$

$\begin{array}{l}⇒r\left(r-1\right)\left(r-2\right)\left(r-3\right)+6r\left(r-1\right)\left(r-3\right)+2r\left(r-1\right)-4r+4=0\\ r\left(r-1\right)\left(r-2\right)\left(r-3\right)+6r\left(r-1\right)\left(r-3\right)+2r\left(r-1\right)-4\left(r-1\right)=0\\ \left(r-1\right)\left(r\left(r-2\right)\left(r-3\right)+6r\left(r-2\right)-2r-4\right)=0\\ \left(r-1\right)\left(r\left(r-2\right)\left(r-3\right)+6r\left(r-2\right)+2\left(r-2\right)\right)=0\\ \left(r-1\right)\left(\left(r-2\right)\left(r\left(r-3\right)+6r+2\right)\right)=0\\ \left(r-1\right)\left(r-2\right)\left(r\left(r-3\right)+6r+2\right)=0\\ \left(r-1\right)\left(r-2\right)\left({r}^{2}-3r+6r+2\right)=0\\ \left(r-1\right)\left(r-2\right)\left({r}^{2}-3r+2\right)=0\\ \left(r-1\right)\left(r-2\right)\left(r+1\right)\left(r+2\right)=0\\ ⇒r=1,r=2,r=-1,r=-2\end{array}$

Therefore,

${y}_{1}={x}^{1},{y}_{2}={x}^{2},{y}_{3}={x}^{-1},{y}_{4}={x}^{-2}$

Are all solution of equation (2)

## Step 3: Solving further:

$\begin{array}{l}{y}_{3}={x}^{2-3i}\\ ={e}^{\left(2-3i\right)lnx}\\ ={x}^{2}\left\{cos\left(-3lnx\right)+isin\left(-3lnx\right)\right\}\\ ={x}^{2}\left\{cos\left(3lnx\right)-isin\left(3lnx\right)\right\}\end{array}$

Hence,

$\left\{{x}^{1},{x}^{2}\left\{cos\left(3lnx\right)+isin\left(3lnx\right)\right\},{x}^{2}\left\{cos\left(3lnx\right)-isin\left(3lnx\right)\right\}$ is the fundamental solution set. ### Want to see more solutions like these? 