### Cauchy-Euler Equations - (3.6) Consider an nth

```Cauchy-Euler Equations - (3.6)
Consider an nth-order nonhomogeneous linear differential equation:
Ly gx where Ly a n x n y n a n"1 x n"1 y n"1 . . . a 1 xy U a 0 y.
Notice that the coefficient functions
a k x a k x k , k 1, . . . , n.
A differential equation in this form is known as a Cauchy-Euler equation. Now let us find the general solution
of a Cauchy-Euler equation.
1. Second Order Homogeneous Cauchy-Euler Equations
Consider the homogeneous differential equation of the form:
a 2 x 2 y UU a 1 xy U a 0 y 0.
Let y x m . Find m so that y is a solution of the equation.
y U mx m"1 , y UU mm " 1 x m"2 .
a 2 x 2 y UU a 1 xy U a 0 y a 2 x 2 mm " 1 x m"2 a 1 xmx m"1 a 0 x m
x m a 2 m 2 " a 2 m a 1 m a 0 x m a 2 m 2 a 1 " a 2 m a 0 0
x p 0 implies
a 2 m 2 a 1 " a 2 m a 0 0.
Let Pm a 2 m 2 a 1 " a 2 m a 0 . So, y x m is a solution of a second order Cauchy-Euler equation if
and only if m is a solution of Pm 0. Since Pm is a second degree polynomial, Pm 0 may have
a. two simple real solutions;
b. one real solution with multiplicity 2; or
c. a pair of conjugate complex solutions.
The table below we list the relation of the zeros of Pm and the solution of the corresponding Cauchy-Euler
equation. Let m 1 and m 2 be real numbers.
solution of Pm 0
y 1 , y 2 " solutions of a Cauchy-Euler equation
y 1 x m1 ,
m m 1 , m 2 where m 1 p m 2
y 1 x m1 ,
m m1, m1
m m 1 im 2 , m m 1 " im 2
y 2 x m2
y 2 x m 1 ln x
y 1 x m 1 cosm 2 ln x , y 2 x m 1 sinm 1 ln x yh c1y1 c2y2
Example Solve the differential equation or the initial value problem
b x 2 y UU " 5 xy " 2y 0
2
2 UU
U
U
c x y " 2xy 2y 0, y1 "1, y 1 1
a x 2 y UU " 3xy U 20y 0
a. x 2 y UU " 3xy U 20y 0, a 2 1, a 1 "3, a 0 20
Pm m 2 "3 " 1 m 20 m 2 " 4m 20 0
4 o 16 " 420 2 o "16 2 o i4
m
2
y 1 x 2 cos4 ln x , y 2 x 2 sin4 ln x ,
The general solution of the differential equation:
1
y h c 1 x 2 cos4 ln x c 2 x 2 sin4 ln x b. x 2 y UU " 5 xy " 2y 0, a 2 1, a 1 " 5 , a 0 "2
2
2
5
7
2
2
Pm m " " 1 m " 2 m " m " 2 1 2m " 7m " 4 1 2m 1 m " 4 0
2
2
2
2
1
m" , m4
2
y 1 x "1/2 1 ,
y2 x4
x
The general solution of the differential equation:
yh c1 1 c2x4
x
c. x 2 y UU " xy U y 0, y1 "1, y U 1 1, a 2 1, a 1 "1, a 0 1
Pm m 2 "1 " 1 m 1 m 2 " 2m 1 m " 1 2 0
m 1, 1
y 1 x, y 2 x ln x
The general solution of the differential equation:
y h c 1 x c 2 x ln x
y Uh c 1 c 2 ln x x 1x
y1 c 1 0 "1
y U 1 c 1 c 2 1
,
1 0
c1
1 1
c2
c 1 c 2 ln x 1 "1
1
,
c 1 "1
c 2 1 " c 1 1 " "1 2
The solution of the initial value problem:
y "x 2x ln x
2. Third-order Homogeneous Cauchy-Euler Equations
Consider:
a 3 x 3 y UUU a 2 x 2 y UU a 1 xy U a 0 y 0
Let y x m . Similarly, we have
y U mx m"1 , y UU mm " 1 x m"2 , y UUU mm " 1 m " 2 x m"3
Then
a 3 x 3 mm " 1 m " 2 x m"3 a 2 x 2 mm " 1 x m"2 a 1 xmx m"1 a 0 x m 0
xm
a 3 m 3 a 2 " 3a 3 m 2 2a 3 " a 2 a 1 m a 0
0
Let
Pm a 3 m 3 a 2 " 3a 3 m 2 2a 3 " a 2 a 1 m a 0 .
So, y x m is a solution of a second order Cauchy-Euler equation if and only if m is a solution of
Pm 0. To solve a third-order Cauchy-Euler equation, we first find all 3 zeros m 1 , m 2 and m 3 of Pm and then find y 1 , y 2 and y 3 based on the types of m Ui s.
Example Solve the Cauchy-Euler equation
2
a x 3 y UUU xy U " y 0
b x 3 y UUU x 2 y UU " 2xy U 2y 0
a. x 3 y UUU xy U " y 0, a 3 1, a 2 0, a 1 1, a 0 "1
Pm m 3 0 " 31 m 2 21 " 0 1 m " 1 m 3 " 3m 2 3m " 1 m " 1 3 0
m 1, 1, 1
y 1 x,
y 2 x ln x,
y 3 xln x 2
y h c 1 x c 2 x ln x c 3 xln x 2
b. x 3 y UUU x 2 y UU " 2xy U 2y 0, a 3 1, a 2 1, a 1 "2, a 0 2
Pm m 3 1 " 3 m 2 2 " 1 " 2 m 2 m 3 " 2m 2 " m 2 m " 1 m " 2 m 1 0
m 1, 2, " 1
y 1 x, y 2 x 2 , y 3 x "1
y h c 1 x c 2 x 2 c 3 1x
3. Second-order and Third-order Nonhomogeneous Cauchy-Euler Equations:
Steps to solve a second-order or third-order nonhomogeneous Cauchy-Euler equation:
a. Solve y h .
b. Use the Method of Variation of Parameters to solve y p .
Example Solve 4x 2 y UU 4xy U " y 12
x .
a. Solve y h from 4x 2 y UU 4xy U " y 0. a 2 4, a 1 4, a 0 "1
Pm 4m 2 4 " 4 m " 1 4m 2 " 1 0
m o1
2
y 1 x 1/2 x,
y 2 x "1/2 1
x
The general solution of the homogeneous differential equation:
yh c1 x c2 1
x
b. Solve y p from y UU 1x y U " 1 2 y 33 .
x
4x
Let y p u 1 x u 2 1 .
x
U
U
Solve u 1 and u 2 from the system:
x 1/2
1
2
3
x "1/2
x "1/2 " 12 x "3/2
u U1
u U2
0
3
x3
u U1
1 "1 1
"2x "
u U2
"x
1
2
1
2
x "1
"3x "7/2
x "3/2
3
x3
" 12 x "1/2 x 1/2
3x "5/2
3x "5/2
0
"x "1/2
" 3x "3/2
Solve u 1 and u 2 :
; u U1 dx ; 3x "5/2 dx "2x "3/2
u 2 ; "3x "3/2 dx 6x "1/2
u1 A particular solution of the nonhomogeneous differential equation:
y p "2x "3/2 x 6x "1/2 1 4x
x
The general solution of the nonhomogeneous differential equation:
y c 1 x c 2 1 4x
x
Example Solve x 3 y UUU x 2 y UU " 2xy U 2y x 3 ln x, given that y h c 1 x c 2 x 2 c 3 1x .
b. Solve y p from y UUU 1x y UU " 22 y U 23 y ln x
x
x
1
2
Let y p u 1 x u 2 x u 3 x .
i. Solve u U1 , u U2 , u U3 from the system:
x x 2 x "1
u U1
1 2x "1 x "2
u U2
0 2
2x
0
u U3
"3
0
ln x
Use Cramer’s Rule to solve u Ui :
x x 2 x "1
det
1 2x "1 x "2
0 2
det
6x ,
det
2x "3
x 2 x "1
0
2x "1 x "2
ln x 2
x 0
x "1
1 0
"1 x "2
2x "3
2x ln x,
det
1 2x 0
0 2
u U2 2x ln x x
6
1 ln x,
3
u U3 x 2 ln x x
6
; " 12 x ln xdx " 14 x 2 ln x 18 x 2
u 2 ; 1 ln x 1 x ln x " 1 x
3
3
3
u 3 ; 1 x 3 ln xdx 1 x 4 ln x " 1 x 4
24
96
6
u1 A particular solution of the nonhomogeneous differential equation:
x 2 ln x
ln x
ii. Solve u 1 , u 2 and u 3 .
4
"3 ln x
x x2 0
0 ln x 2x "3
u U1 "3 6ln x " 1 x ln x,
2
x
0
1 x 3 ln x
6
1 x ln x " 1 x x 2 1 x 4 ln x " 1 x 4
" 1 x 2 ln x 1 x 2 x 8
96
4
3
3
24
1
3
x 4 ln x " 7 32
The general solution of the nonhomogeneous differential equation:
y c 1 x c 2 x 2 c 3 1x 1 x 3 4 ln x " 7 32
yp 5
1
x
```