Question

Higher-Order Cauchy–Euler Equations. A differential equation that can be expressed in the form

$a_n{x}^n{y}^{\left(n\right)}\left(x\right)+a_{n-1}{x}^{n-1}{y}^{(n-1)}\left(x\right)+...+a_{0}y\left(x\right)=0$

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) $x^{3}y\text{'}\text{'}\text{'}+x^{2}y\text{'}\text{'}-2xy\text{'}+2y=0,x>0$

(b) $x^4{y}^{\left(4\right)}+6x^{3}y\text{'}\text{'}\text{'}+2x^{2}y\text{'}\text{'}-4xy+4y=0,x>0$

(c) $x^{3}y\text{'}\text{'}\text{'}-2x^{2}y\text{'}\text{'}+13xy\text{'}-13y=0,x>0$

[Hint: $x^{\alpha +\beta i}=e^{(\alpha +\beta i)lnx}=x^a\left\{cos\right(\beta lnx)+isin(\beta lnx\left)\right\}$]

Short answer

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

Step-by-step solution

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,

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

Substitution in equation (1) we get,

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

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

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

Therefore,

$y_1=x^1,y_2=x^2,y_3=x^{-1}$

Are all solutions of equation (1)

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

Solution for (b):

(b)Given differential equation is,

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

Let,

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

Substituting in equation (2), we get

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

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

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

Therefore,

$y_1=x^1,y_2=x^2,y_3=x^{-1},y_4=x^{-2}$

Are all solution of equation (2)

Solving further:

$\begin{array}{l}{y}_3=x^{2-3i}\\ =e^{(2-3i)lnx}\\ =x^2\left\{cos\right(-3lnx)+isin(-3lnx\left)\right\}\\ =x^2\left\{cos\right(3lnx)-isin(3lnx\left)\right\}\end{array}$

Hence,

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