2 Lecture Notes on Cauchy-Euler DEs

Math 252 – <strong>Lecture</strong> <strong>Notes</strong> #2
David Maslanka
REDUCTION OF A CAUCHY–EULER DE TO CONSTANT COEFFICIENTS
Consider the fourth order homogeneous Cauchy–Euler DE:
4
3
4 d y 3 d y
ax +
4
3
dx
2 d y
bx + cx +
2
dx dx
2
dy
dx +ε y = 0.
dx
( 1 )
We may transform ( 1 ) into a constant coefficient DE with the substitution:
t
x = e ⇔ t = ln x for x ∈( 0 , + ∞ ) .
Now we have the representations for the first four derivatives of y with respect to x:
dy =
dx
dy
• = ( ln x)
dt
dt
dx
dy d
• =
dt dx
1 dy
• by the Chain Rule.
x dt
2
d y
2
dx
2
d y
2
dx
=
1 d ⎛ dy ⎞ dt
• ⎜ ⎟ •
x dt ⎝ dt ⎠ dx
2
1 d y ⎛ 1
x • dt
⎟ ⎞
• ⎜
⎝ x ⎠
=
2
dy d ⎛ 1 ⎞
+ • ⎜ ⎟ by the Product and Chain Rules
dt dx ⎝ x ⎠
dy
+ •
dt
1
⎛
⎜ − 2
⎝ x
⎞
⎟
⎠
= 1 ⎛
2
2 2 ⎟ ⎞
⎜
d y dy
•
−
x ⎝ dt dt ⎠
3
d y
3
dx
3
d y
3
dx
= 1 2
d ⎛d
y dy
2 2 ⎟ ⎞
• ⎜
−
x dt ⎝ dt dt ⎠
dt
• +
dx
2
⎛d
y
⎜
2
⎝ dt
dy⎞
− ⎟ •
dt
⎠
= 1 3 2
⎛d
y d y
2 3 2 ⎟ ⎞
• ⎜
⎛1⎞
− • ⎜ ⎟⎠
x ⎝ dt dt ⎠ ⎝x + 2
⎛ d y dy
⎟ ⎞
⎜
− •
2
⎝ dt dt ⎠
d
dx
⎛
⎜
⎝
1
2
x
⎞
⎟
⎠
⎛ − 2 ⎞ 1
⎜
3
⎟ =
⎝ x ⎠ x
by the Product and Chain Rules
3 2
⎛ d y d y
• ⎜
− 3
3 3 2
⎝ dt dt
dy⎞
+ 2 ⎟
dt
⎠
4
d y
4
dx
4
d y
4
dx
4
d y
4
dx
3
2
3
2
1 d ⎛d
y d y dy⎞
dt ⎛d
y d y dy⎞
d
= • ⎜
⎟
− 3 + 2 •
3 3
2
x dt
+ ⎜
⎟
⎛ 1 ⎞
⎝ dt dt dt ⎠ dx
− 3 + 2 •
3
2
⎜
3
⎟
⎝ dt dt dt ⎠ dx ⎝ x ⎠
–by the Product and Chain Rules
4
3
2
3
2
1 ⎛d
y
⎞
= • ⎜
⎟ •
3
x
− d y d y ⎛ 1 ⎞ ⎛ d y d y dy⎞
3 + 2
4
3
⎜ ⎟ + ⎜
⎟
⎛ − 3 ⎞
⎝ dt dt dt
2
⎠ ⎝ x ⎠
− 3 + 2 •
3
2
⎜
4
⎟
⎝ dt dt dt ⎠ ⎝ x ⎠
1
=
4
x
4
3
2
⎛ d y d y
• ⎜
−
⎝ dt
6 4
3
dt
2
d y
+ 11
dt
dy
⎟ ⎞
− 6
dt ⎠
dy d y
Upon substitution for , dx
2
dx
2
3
d y
dx
,
3
4
d y
and
4
dx
back into (1) we find that
4
⎛d
y
a ⎜
4
⎝ dt
−
3
2
6 2
d y d y
+ 11
3
dt dt
− 6
dy⎞
⎟
dt
⎠
+ b
3
⎛d
y
⎜
−
3
⎝ dt
d
2
dt
y
3
2
dy⎞
+ 2 ⎟
dt
⎠
2
⎛d
y dy⎞
+ c ⎜ ⎟
−
2
⎝ dt dt ⎠
dy
+ d + ε y = 0 .
dt
1

Regrouping terms, we have the equivalent constant coefficient DE in terms of the
independent variable t:
4
d y
dt
a
4
d y
+ (–6a + b)
3
dt
3
d y
+ (11a – 3b + c )
2
dt
2
dy
+ (–6a + 2b – c + d ) + ε y = 0
dt
( 2 )
Now to solve for y in ( 2 ) we set y =
[ a
4
m + (–6a + b)
3
m + (11a – 3b + c )
mt
e which yields:
2
m + (–6a + 2b – c + d ) m + ε y ]
Consequently, the characteristic equation corresponding to ( 2 ) is:
mt
e = 0 .
4
a m + (–6a + b)
3
m + (11a – 3b + c )
2
m + (–6a + 2b – c + d ) m + ε y = 0
( * )
We determine several solutions for ( 1 ) based on the roots to ( * ).
Case 1: Distinct Real Roots
In case ( * ) factors as: ( m ‐ m1)
( m ‐ m1)
( m ‐ m1)
( m ‐ m1)
= 0
where m i 2 R for i = 1, 2, 3, 4 then
m = m
1
, m
2
, m
3
, m
4
are all distinct real roots to ( * ). So
m1
t m2
t m3
t m4
t
y = c1
e + c2
e + c3
e + c4
e
ξ ln x ξ
is the solution to ( 2 ). Since t = ln x and e = x we find
m1
m2
m3
m4
y = c x + c x + c x + c x
to be the general solution to ( 1 ) for ∈( 0 , + ∞ )
x .
Case 2: A Repeated Real Root of Multiplicity Four
4
In case ( * ) factors as: ( m − m 0 ) = 0
then m = m
0
is a root of multiplicty four to ( * ). So
m0
t m0
t 2 m0
t 3 m0
t
y = c1
e + c2
t e + c3
t e + c4
t e
is the solution to ( 2 ). Upon substituting t = ln x we find
to be the general solution to ( 1 ) for ∈( 0 , + ∞ )
1
x .
Case 3: A Repeated Pair of Conjugate Roots
2
2
In case ( * ) factors as: ( m − ( α + i β )) ( m − ( α − i β )) = 0
then m = α ± i β is a repeated pair of conjugate roots to ( * ). So
t
y = e α t
( c1
+ c2t
) cos( β t ) + e α ( c3
+ c4t
) sin( β t )
is the solution to ( 2 ). Then setting t = ln x we find
2
m0
m0
2 m0
3 m0
y = c x + c ( ln x)
x + c ( ln x)
x + c ( ln x)
x
1
2
3
3
4
4
2

y =
α
x ( c1
c2
+ ln x ) cos ( β ln x ) +
α
x ( c3
c4
+ ln x ) sin ( β ln x )
to be the general solution to ( 1 ) for x ∈( 0 , + ∞ ) .
It is important to recognize that upon setting y =
m
x in ( 1 ) yields:
[ a m ( m – 1 )( m – 2 )( m – 3 ) + b m ( m – 1 )( m – 2 ) + c m ( m – 1 ) + d m + ε ] x m = 0
That is,
4
3
2
m
[ a m + (–6a + b) m + (11a – 3b + c) m + (–6a + 2b – c + d ) m + ε y ] x = 0 .
Thus, y = x m is a solution to ( 1 ) ⇔ m is a root of:
a 4
3
2
m + (–6a + b) m + (11a – 3b + c ) m + (–6a + 2b – c + d ) m + ε y = 0
( 3 )
Observe that ( 3 ) agrees precisely with the characteristic equation ( * ) for ( 2 ) .
o-------------o-------------o-------------o-------------o-------------o-------------o-------------o
In general, the
th
n order C-E DE:
n
n -1
n d y
n -1 d y
a n x +
n n - 1 x
n -1
dx
a + . . . +
dx
dy
a 1 x + a 0 y = 0
dx
( 4 )
when transformed to the constant coefficient DE:
n
n -1
d y d y
dy
A + n A
n n -1
+ . . . + A
n -1
1 + A 0 y = 0
dt dt
dt
by setting x = e
t will possess the characteristic equation:
α n
n m + n -1
α n - 1 m + . . . + α 1 m + α 0 = 0
if and only if the equation having precisely the same coefficients for m:
n
[ α n m + n -1
α n - 1 m + . . . + α 1 m + α 0
]
m
x = 0
is obtained directly from ( 4 ) by setting y = x m .
3