LECTURE 2: CAUCHY-EULER EQUATIONS
MINGFENG ZHAO
July 08, 2015
The Cauchy-Euler Equations
A Cauchy-Euler equation has the form of:
x2 y 00 + αxy 0 + βy = 0,
(1)
where α, β are real constants.
Let x = et (if we assume x > 0), by the chain rule, we have
dy
dt
=
=
d2 y
dt2
=
=
=
dy dx
dy
·
= et ·
dx dt
dx
dy
x·
dx
dy
d2 y dx
et ·
+ et · 2 ·
dx
dx
dt
2
dy
d y
et ·
+ e2t · 2
dx
dx
d2 y
dy
+ x2 · 2 .
x·
dx
dx
By the product rule
That is, we have
x·
dy
dy
=
,
dx
dt
and x2 ·
d2 y
d2 y
dy
d2 y dy
= 2 −x·
= 2 −
.
2
dx
dt
dx
dt
dt
By (1), we have
d2 y dy
dy
−
+α·
+ βy = 0.
2
dt
dt
dt
Hence, to solve (1), it’s equivalent to solve the following second order constant coefficients linear differential equation:
(2)
d2 y
dy
+ (α − 1) ·
+ βy = 0.
dt2
dt
A solution to (2) has the form of y = ert , where r is a solution to r2 + (α − 1)r + β = 0 (r may be complex number),
since x = et , then t = ln x and a solution to (1) has the form of y = xr .
For solution to (2), we have three cases, then they correspond three cases for solutions to (1):
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MINGFENG ZHAO
• If (α − 1)2 − 4β > 0, that is, r2 + (α − 1)r + β = 0 has two different real roots r1 and r2 , then the solutions to
(2) have the form of y = C1 er1 t + C2 er2 t , which implies that the solutions to (1) have the form of
y = C1 xr1 + C2 xr2 .
• If (α − 1)2 − 4β = 0, that is, r2 + (α − 1)r + β = 0 has only one real root r0 , then the solutions to (2) have the
form of y = C1 er0 t + C2 ter0 t , which implies that the solutions to (1) have the form of
y = C1 xr0 + C2 xr0 ln x.
• If (α − 1)2 − 4β < 0, that is, r2 + (α − 1)r + β = 0 has two complex roots r1 = a + ib and r2 = a − ib, then the
solutions to (2) have the form of y = C1 eat cos(bt) + C2 eat sin(bt), which implies that the solutions to (1) have
the form of
y = C1 xa cos(b ln x) + C2 xa sin(b ln x).
Remark 1. For the case that (α − 1)2 − 4β = 0, that is, r2 + (α − 1)r + β = 0 has only one real root, say r0 , then
r2 + (α − 1)r + β = (r − r0 )2 = r2 − 2r0 r + r02 , which implies that α − 1 = −2r0 and β = r02 . There are another two
approaches to find another solution to (1):
• Approach I: (Method of reduction of order) Let u(x) = C(x)xr0 be a solution to (1) for some function C(x),
let’s determine what C(x) is, plug u(x) into (1), then
u0 (x)
=
C 0 (x)xr0 + r0 C(x)xr0 −1
u00 (x)
=
C 00 (x)xr0 + r0 C 0 (x)xr0 −1 + r0 C 0 (x)xr0 −1 + r0 (r0 − 1)C(x)xr0 −2
= C 00 (x)xr0 + 2r0 C 0 (x)xr0 −1 + r0 (r0 − 1)C(x)xr0 −2 .
So we get
0
= x2 u00 (x) + αxu0 (x) + βu(x)
= C 00 (x)xr0 +2 + 2r0 C 0 (x)xr0 +1 + r0 (r0 − 1)C(x)xr0 + αC 0 (x)xr0 +1 + αr0 C(x)xr0 + βC(x)xr0
= C 00 (x)xr0 +2 + (2r0 + α)C 0 (x)xr0 +1 + [r0 (r0 − 1) + αr0 + β]C(x)xr0
= C 00 (x)xr0 +2 + C 0 (x)xr0 +1
Since α − 1 = −2r0 and r0 is a root of r(r − 1) + αr + β = 0.
So we get xC 00 (x) + C 0 (x) = 0, then C 0 (x) =
u(x) = xr0 ln x is a solution to (1).
D
, which implies that C(x) = D ln x + E. Hence we know that
x
LECTURE 2: CAUCHY-EULER EQUATIONS
3
• Approach II: (Frobenius’ Method) Think r is a variable, let φ(r, x) = xr , then
x2 φxx + αxφx + βφ = r(r − 1)xr + rαxr + βxr
= xr [r(r − 1) + αr + β]
= xr (r − r0 )2
Since r0 is the unique solution to r(r − 1) + αr + β = 0.
Then we get
x2 φxxr + φxr + βφr = xr ln x(r − r0 )2 + 2xr (r − r0 ).
Then we get
x2 φxxr (x, r0 ) + αxφxr (x, r0 ) + βφr (x, r0 ) = 0.
That is, y = φr (x, r0 ) is a solution to (1). Notice that
φr (x, r0 ) = xr0 ln x.
Remark 2. For the case that (α − 1)2 − 4β < 0, we can also use the form C1 xr1 + C2 xr2 , in fact, let r1 = a + ib, by
Euler’s identity, we have
xa+ib
= xa · xib
= xa · eib ln x
= xa [cos(b ln x) + i sin(b ln x)]
= xa cos(b ln x) + ixa sin(b ln x).
So the real and imaginary parts or xr1 form a fundamental set of solutions to (1).
Definition 1. For the Cauchy-Euler equation x2 y 00 + αxy 0 + βy = 0, the polynomial r(r − 1) + αr + β is called the
indicial polynomial for this Cauchy-Euler equation.
Remark 3. A general n-th order Cauchy-Euler equation has the form:
xn y (n) + α1 xn−1 y (n−1) + · · · + αn−1 xy 0 + αn y = 0.
Example 1. Find the general solution to x2 y 00 + 2xy 0 − 2y = 0 for x > 0.
Try y = xr to be a solution of x2 y 00 + 2xy 0 − 2y = 0, then
y0
= rxr−1
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MINGFENG ZHAO
y 00
0
= r(r − 1)xr−2
= x2 y 00 + 2xy 0 − 2y
= r(r − 1)xr + 2rxr − 2xr
= xr [r(r − 1) + 2r − 2].
Then r(r − 1) + 2r − 2 = 0, that is, r satisfies r2 + r − 2 = 0. Then we get
r1 = 1,
and r2 = −2.
Hence the general solution to x2 y 00 + 2xy 0 − 2y = 0 is given by:
y = C1 x + C2 x−2 .
Example 2. Find the general solution to x2 y 00 + 3xy 0 + y = 0 for x > 0.
Try y = xr to be a solution of x2 y 00 + 3xy 0 + y = 0, then
y0
= rxr−1
y 00
= r(r − 1)xr−2
0
= x2 y 00 + 3xy 0 + y
= r(r − 1)xr + 3rxr + xr
= xr [r(r − 1) + 3r + 1].
Then r(r − 1) + 3r + 1 = 0, that is, r satisfies r2 + 2r + 1 = 0. Then we get
r1 = r2 = −1.
Hence the general solution to x2 y 00 + 3xy 0 + y = 0 is given by:
y = C1 x−1 + C2 x−1 ln x.
Example 3. Find the general solution to x2 y 00 − xy 0 + 2y = 0 for x > 0.
Try y = xr to be a solution of x2 y 00 − xy 0 + 2y = 0, then
y0
= rxr−1
y 00
= r(r − 1)xr−2
LECTURE 2: CAUCHY-EULER EQUATIONS
0
5
= x2 y 00 − xy 0 + 2y
= r(r − 1)xr − rxr + 2xr
= xr [r(r − 1) − r + 2].
Then r(r − 1) − r + 2 = 0, that is, r satisfies r2 − 2r + 2 = 0. Then we get
r1 = 1 + i,
and r2 = 1 − i.
Then by the Euler’s identity, we have
x1+i
= e(1+i) ln x = eln x+i ln x
= eln x [cos(ln x) + i sin(ln x)]
= x cos(ln x) + ix sin(ln x).
Hence the general solution to x2 y 00 − xy 0 + 2y = 0 is given by:
y = C1 x cos(ln x) + C2 x sin(ln x).
Example 4. Find the general solution to (x − 1)y 00 + y 0 = 0 for x > 1.
It’s equivalent to find the general solution to (x − 1)2 y 00 + (x − 1)y 0 = 0. Try y = (x − 1)r to be a solution of
(x − 1)2 y 00 + (x − 1)y 0 = 0, then
y0
= rxr−1
y 00
= r(r − 1)xr−2
0
=
(x − 1)2 y 00 + (x − 1)y 0
= r(r − 1)(x − 1)r + r(x − 1)r
=
(x − 1)r [r(r − 1) + r].
Then r(r − 1) + r = 0, that is, r satisfies r2 = 0. Then we get
r1 = r2 = 0.
Hence the general solution to (x − 1)y 00 + y 0 = 0 is given by:
y = C1 + C2 ln(x − 1).
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MINGFENG ZHAO
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Example 5. Find the solution to x2 y 00 − xy 0 + y = 0, y(−1) = 0 and y 0 (−1) = 1.
4
3
2 00
0
r
Try y = (−x) to be a solution of x y − xy + y = 0, then
4
y0
= −r(−x)r−1
y 00
= r(r − 1)(−x)r−2
0
3
= x2 y 00 − xy 0 + y
4
3
= r(r − 1)xr − r(−x)r + (−x)r
4
3
= (−x)r r(r − 1) − r +
.
4
3
3
= 0, that is, r satisfies r2 − 2r + = 0. Then we get
4
4
1
3
r1 = , and r2 = .
2
2
3
Hence the general solution to x2 y 00 − xy 0 + y = 0 is given by:
4
Then r(r − 1) − r +
1
3
y = C1 (−x) 2 + C2 (−x) 2 .
Then we get
y 0 (x) = −
1
3
C1
3C2
(−x) 2 −
(−x) 2 .
2
2
Since y(−1) = 0 and y 0 (−1) = 1, then
C1
3C2
−
= 1.
2
2
3
So we get C1 = 1 and C2 = −1. Therefore, the solution to x2 y 00 − xy 0 + y = 0, y(−1) = 0 and y 0 (−1) = 1 is given
4
by:
C1 + C2 = 0,
and
1
−
3
y = (−x) 2 − (−x) 2 .
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca