Math 201 Lecture 12: Cauchy-Euler Equations
Feb. 3, 2012
•
Many examples here are taken from the textbook. The first number in () refers to the problem number
in the UA Custom edition, the second number in () refers to the problem number in the 8th edition.
0. Review
•
To solve general 2nd order linear equations,
a(t) y ′′ + b(t) y ′ + c(t) y = f (t).
(1)
1. Guess one solution y1. (Popular guesses: constants; exponentials; simple polynomials; sin ,cos)
2. Write the equation into standard form
y ′′ + p(t) y ′ + q(t) y = 0
(2)
and apply the “reduction of order” formula:
Z −R p(t)dt
e
y2(t) = y1(t)
dt
y1(t)2
(3)
3. Apply variation of parameters formula
y p = v1 y1 + v2 y2
(4)
with
v1 =
Z
−f (t) y2(t)
,
a(t) [y1 y2′ − y1′ y2]
v2 =
Z
f (t) y1(t)
.
a(t) [y1 y2′ − y1′ y2]
(5)
to obtain yp. Note that a(t) is not a constant anymore.
4. The general solution is then given by
y = C1 y1 + C2 y2 + y p.
(6)
Simplify if possible.
5. Check your solution if time allows.
•
Quiz: Solve
t y ′′ − y ′ = 1.
(7)
(As t = 0 is a singular point, let’s just consider t > 0).
Solution.
1. Guess one solution for
t y ′′ − y ′ = 0.
(8)
It is clear that any constant is a solution. So let’s take y1 = 1.
2. Use reduction of order to get y2.
a. Write the equation in standard form:
y ′′ −
1 ′ 1
y =
t
t
b. Compute
y2 = y1
Z
R
e− p
=
y12
Z
3. Use variation of parameters to get y p.
1
e
R
1
t
1
p(t) = − .
t
=
Z
eln t =
Z
(9)
t=
t2
.
2
(10)
Math 201 Lecture 12: Cauchy-Euler Equations
2
Compute
2 ′
t
t2
′
y2] = t 1
− (1) = t2.
2
2
Z
Z
−f (t) y2(t)
t
1 · t2
v1 =
=−
=− ;
′
′
a(t) [y1 y2 − y1 y2]
2
2 t2
Z
Z
1
1·1
f (t) y1(t)
=
=−
v2 =
t
t2
a(t) [y1 y2′ − y1′ y2]
So
t
1 t2
yp = − 1 + −
= −t.
2
t 2
4. The general solution is
t2
y = C1 + C2 − t
2
which can be simplifiied to
a(t) [y1 y2′ − y1′
y = C1 + C2 t2 − t.
(11)
(12)
(13)
(14)
(15)
(16)
1. Basic Information
•
The Equation:
a t2 y ′′ + b t y ′ + c y = 0.
(17)
Such equations are called “Cauchy-Euler equations” and they are as easy as equations of constant
coefficients (In fact, Cauchy-Euler equations is just constant-coefficient equations in disguise, see Notes
and Comments).
•
How to get general solution
◦
Idea: From general theory of 2nd order linear equations, we know that as soon as we figure out
two linearly independent solutions y1, y2, the general solution is simply
C1 y1 + C2 y2.
(18)
Recall that in studying constant-coefficient equations a y ′′ + b y ′ + c y = 0 we obtain y1, y2
through guess y = ert. Here the idea is similar but the guess is different: y = tr.
◦
Procedure:
1. Write down the characteristic equation
a r 2 + (b − a) r + c = 0.
(19)
y1 = tr1, y2 = tr2;
(20)
y1 = tr , y2 = tr ln t.
(21)
Solve it to get r1, r2.
2. Three cases:
−
−
−
r1 r2, both real:
r1 = r2 = r, real.
r1,2 = α ± i β.
y1 = tα cos (β ln t),
y2 = tα sin (β ln t).
(22)
3. Write down general solution.
◦
Examples:
Example 1. Solve
t2 y ′′ + 7 t y ′ − 7 y = 0.
(23)
Feb. 3, 2012
3
Solution. Guided by the theory, we only need to find two linearly independent solutions. The
key now is to realize the following property of tr: (tr)(k) tk = C tr. Substitute y = tr into the
equation, we have
0 = t2 y ′′ + 7 t y ′ − 7 y = [r (r − 1) + 7 r − 7] tr = 0
r2 + 6 r − 7 = 0
r1 = −7, r2 = 1.
(24)
Thus the general solution is
y = c1 t−7 + c2 t.
(25)
t2 y ′′ − 3 t y ′ + 4 y = 0.
(26)
Example 2. solve
Solution. Substituting y = tr we have
r (r − 1) − 3 r + 4 = 0
r2 − 4 r + 4 = 0
r1 = r2 = 2.
(27)
This is double root so the general solution is given by
y = c1 t2 + c2 t2 ln t.
(28)
1 ′ 5
y + 2 y = 0.
t
t
(29)
t2 y ′′ − t y ′ + 5 y = 0.
(30)
Example 3. Solve
y ′′ −
Solution. Multiply both sides by t2:
Substituting
y = tr
gives
r (r − 1) − r + 5 = 0
r2 − 2 r + 5 = 0
r1 = 1 + 2 i, r2 = 1 − 2 i.
(31)
The general solution is then given by
y = c1 t cos (2 ln t) + c2 t sin (2 ln t).
•
(32)
Relations to constant-coefficient equations. The formulas
◦
◦
◦
r1 r2, both real:
y1 = tr1, y2 = tr2;
(33)
y1 = tr , y2 = tr ln t.
(34)
r1 = r2 = r, real.
r1,2 = α ± i β.
y1 = tα cos (β ln t),
y2 = tα sin (β ln t).
(35)
above looks similar to our theory for linear constant-coeffcient equations. This similarity becomes
more striking if we introduce a new variable x = ln t: The three cases become
◦
Two distinct roots
◦
One double root
◦
Complex roots
er1 x , er2 x;
erx , x erx;
eαx cos β x, eαx sin β x.
This is no coincidence! In fact, setting x = ln t gives
dy
d2 y
d −1 dy dx
d2 y
dy
dy dy dx
=
= t−1
, y′ = 2 =
= t−2 2 − t−2 .
y′ =
t
dx
dt
dx
dx
dx
dt dx dt
dx dt
(36)
Substituting into the equation
0 = a t2 y ′′ + b t y ′ + c y = a
d2 y
dy
+ (b − a)
+ c y.
2
dx
dx
(37)
Math 201 Lecture 12: Cauchy-Euler Equations
4
Thus we have transformed the Euler-Cauchy equation into a constant-coefficient equaiton. Furthermore, the auxiliary equation for this equation is
a r 2 + (b − a) r + c = a r (r − 1) + b r + c
(38)
which is exactly the characteristic equation of the Cauchy-Euler equation!
•
How to check solutions
◦
Note: Check solutions, especially in the 3rd case, may involve so much calculation that it
becomes not worthwhile. Instead, make sure you write down the correct characteristic equation
and solve it correctly.
2. Things to be Careful/Tricky Issues
See “Common Mistakes” for examples.
•
Fail to see that the equation is Cauchy-Euler. (This is the hottest mistake in 201!!)
•
After finding r1,2, write er1 t , er2 t instead of tr1, tr2. (This mistake is also very popular.)
3. More Examples
•
Initial value problem.
Example 4. Solve the following initial value problem for the Cauchy-Euler equation
t2 y ′′ − 4 t y ′ + 4 y = 0;
Solution. Substituting y = tr gives
r (r − 1) − 4 r + 4 = 0
y(1) = −2,
r2 − 5 r + 4 = 0
y ′(1) = −11.
r1 = 4,
r2 = 1.
(39)
(40)
Thus the general solution is given by
y(t) = c1 t4 + c2 t.
(41)
Using the initial values, we have
− 11 = y ′(1) = 4 c1 + c2.
−2 = y(1) = c1 + c2;
(42)
Solving this we reach
c1 = −3,
c2 = 1.
(43)
Thus the solution to the initial value problem is given by
y(t) = −3 t4 + t.
•
(44)
What happens if t < 0?
Example 5. (NA; 4.7 15) Solve for t < 0
y ′′ −
1 ′ 5
y + 2 = 0.
t
t
Solution. Let x = −t. Then x > 0. We have
dy
d2 y d dy
dy dy dx
d dy
d dy
dx d2 y
=
=− ;
=
=
.
=
−
=
−
·
2
dx
dt
dt dt
dt dx dt
dt dx
dx dx
dt dx2
(45)
(46)
So if we use x instead of t as the variable, the equation (with unknown y and variable x) reads
d2 y 1 dy
5
−
+
= 0.
dx2 x dx x2
(47)
It is still Cauchy-Euler, with a = 1, b = −1, c = 5. We write down characteristic equation
r2 − 2 r + 5 = 0
r1,2 = 1 ± 2 i.
(48)
Feb. 3, 2012
5
So the solution reads
y(x) = C1 x cos (2 ln (x)) + C2 x sin (2 ln x).
(49)
y(t) = C1 (−t) cos (2 ln (−t)) + C2 (−t) sin (2 ln (−t)).
(50)
Back to t:
Which can be written as
y(t) = C1 |t| cos (2 ln |t|) + C2 |t| sin (2 ln |t|).
(51)
Remark 6. In fact, we can solve Cauchy-Euler for t 0 as
◦
◦
◦
r1 r2, both real:
y1 = |t|r1, y2 = |t|r2;
(52)
y1 = |t|r , y2 = |t|r ln |t|.
(53)
r1 = r2 = r, real.
r1,2 = α ± i β.
y1 = |t|α cos (β ln |t|),
•
y2 = |t|α sin (β ln |t|).
(54)
What if it’s not t?
Example 7. (4.7 21; 4.7 21) Solve
(t − 2)2 y ′′ − 7 (t − 2) y ′ + 7 y = 0,
t > 2.
(55)
Solution. It is clear that we should introduce x = t − 2. Now chain rule gives
dy dy
= ;
dt dx
The equation becomes
x2
This can be easily solved:
d2 y d2 y
=
.
dt2 dx2
dy
d2 y
−7x
+ 7 y = 0,
dx
dx2
x > 0.
(56)
(57)
y(x) = C1 x + C2 x7.
(58)
y(t) = C1 (t − 2) + C2 (t − 2)7.
(59)
Back to t:
•
Nonhomogeneous problem.
◦
There are two ways to attack nonhomogeneous problem for Cauchy-Euler equations.
1. Introduce x = ln t, transform it to constant-coefficient case, then apply undetermined
coefficients, or variation of parameters;
2. Apply variation of parameters directly.
◦
Examples.
Example 8. Solve
t2 y ′′ − 4 t y ′ + 4 y = t2.
(60)
Solution 1 (x = ln t).
We know that setting x = ln t transforms
a t2 y ′′ + b t y ′ + c y to a
So the equation for x is (x = ln t
d2 y
dy
+ (b − a)
+ c y.
2
dx
dx
(61)
t = ex):
d2 y
dy
−5
+ 4 y = e2x.
2
dx
dx
(62)
Math 201 Lecture 12: Cauchy-Euler Equations
6
We see that this equation is eligible for undetermined coefficients (keep in mind that whenever
undetermined coefficients applys, it is more efficient than variation of parameters).
To solve it we first get y1, y2 by solving
d2 y
dy
−5
+4y=0
dx
dx2
which gives
(63)
y1 = e4x , y2 = ex.
r1 = 4, r2 = 1;
(64)
Now guess
yp = A xs e2x.
(65)
We have s = 0 because 2 does not appear in the root list r1 = 4, r2 = 1.
Substitute yp = A e2x into the equation we get
d2 y
dy
= 4 A e2x ,
= 2 A e2x
dx2
dx
So y p = −
e2x
.
2
4 A − 10 A + 4 A = 1
1
A=− .
2
(66)
The general solution (in x) is then
y = C1 e4x + C2 ex −
1 2x
e .
2
(67)
1 2
t.
2
(68)
Back to t: (Replace every x by ln t):
y(t) = C1 t4 + C2 t −
Solution 2 (Direct application of variation of parameters).
The equation is Cauchy Euler so we can solve the homogeneous equation
t2 y ′′ − 4 t y ′ + 4 y = 0
as follows:
r (r − 1) − 4 r + 4 = 0
So
r2 − 5 r + 4 = 0
(69)
r1 = 4, r2 = 1.
(70)
y1 = t4, y2 = t.
(71)
a(t) [y1 y2′ − y1′ y2] = t2 [t4 − 4 t3 t] = −3 t6.
(72)
Now calculate:
Thus
Z
Z
1
−t2 t
1
−f (t) y2(t)
=
=
= − t−2.
6
−3 t6
3 t3
a(t) [y1 y2′ − y1′ y2]
Z
Z
f (t) y1(t)
t
t2 t4
v2 =
=
=− .
′
′
a(t) [y1 y2 − y1 y2]
3
−3 t6
t2
1 −2 4
t
yp = v1 y1 + v2 y2 = − t
t + − t=− .
2
6
3
v1 =
Therefore
Z
(73)
(74)
(75)
The general solution is then given by
y(t) = C1 t4 + C2 t −
1 2
t.
2
(76)
Example 9. (4.7 41; 4.7 41) Solve
t2 z ′′ + t z ′ + 9 z = −tan (3 ln t).
(77)
Solution. First solve the homogeneous equation
t2 z ′′ + t z ′ + 9 z = 0.
(78)
Feb. 3, 2012
7
It is Cauchy-Euler. So first solve the characteristic equation:
r (r − 1) + r + 9 = 0
So we have
y1 = cos (3 ln t),
r1,2 = ±3.
(79)
y2 = sin (3 ln t).
(80)
Next we use variation of parameters to obtain yp. First calculate:
3
3
′
′
2
a [y1 y2 − y1 y2] = t cos (3 ln t)
sin (3 ln t) − − cos (3 ln t) sin (3 ln t)
t
t
3
= t2 = 3 t.
t
Now we have
Z
−tan (3 ln t) sin (3 ln t)
v1 = −
dt
3t
Z
1
sin2 (3 ln t)
dln t.
=
3
cos (3 ln t)
Letting x = ln t, we compute
Z
Z
Z
sin2 (3 x)
dx
dx =
−
cos (3 x) dx
cos (3 x)
Z cos (3 x)
1
1
cos (3 x) dx − sin (3 x)
=
2 (3 x)
3
cos
Z
1
1
1
=
dsin (3 x) − sin (3 x)
3 1 − sin2 (3 x)
3
Z
Z
dsin (3 x)
1
1
dsin (3 x)
+
− sin (3 x)
=
1 + sin (3 x)
6
3
1 − sin (3 x)
1
1
= [ln |1 + sin (3 x)| − ln |1 − sin (3 x)|] − sin (3 x).
6
3
Back to t:
1 1
1
v1 =
[ln |1 + sin (3 ln t)| − ln |1 − sin (3 ln t)|] − sin (3 ln t) .
3 6
3
(81)
(82)
(83)
(84)
On the other hand
Z
(−tan (3 ln t)) cos (3 ln t)
dt
3t
Z
sin (3 ln t)
= −
dt
3t
Z
1
sin (3 ln t) dln t
= −
3
1
=
cos (3 ln t);
9
Putting things together we have
v2 =
1
[ln |1 + sin (3 ln t)| − ln |1 − sin (3 ln t)|] cos (3 ln t)
18
1
− sin (3 ln t) cos (3 ln t)
9
1
+ cos (3 ln t) sin (3 ln t)
9
1
[ln |1 + sin (3 ln t)| − ln |1 − sin (3 ln t)|] cos (3 ln t)
=
18
Thus the solution is
(85)
yp =
y = C1 cos (3 ln t) + C2 sin (3 ln t) +
1
[ln |1 + sin (3 ln t)| − ln |1 − sin (3 ln t)|] cos (3 ln t).
18
(86)
(87)
Math 201 Lecture 12: Cauchy-Euler Equations
8
Remark 10. For such problems sometimes it is more efficient to let x = ln t and transform the equation
(See Notes and Comments). For example for the above problem, if we let x = ln t, then the equation
becomes
d2 y
+ 9 y = −tan (3 x).
(88)
dx2
This can be solved by variation of parameters, to obtain
y = C1 cos (3 x) + C2 sin (3 x) +
1
[ln |1 + sin (3 x)| − ln |1 − sin (3 x)|] cos (3 x).
18
(89)
Replace x by ln t we get our solution.
4. Notes and Comments
•
Why b − a?
To remember it, just remember that our guess is y = tr. Substituting this into the equation we reach
a r (r − 1) + b r + c = 0
(90)
a r 2 + (b − a) r + c = 0.
(91)
which is exactly
•
Why do we need t > 0?
Note that, unlike ert, tr is singular (meaning: either it’s infinity, or its certain order of derivative
is infinity) at t = 0.
More accurately, when writing the Cauchy-Euler equation in standard form
y ′′ +
b/a
c/a
b/a ′ c/a
y + 2 y = 0.
t
t
(92)
We see that p(t) = t and q(t) = t2 are singular at 0. This is an indication that good theories
(solution exists, solution is unique, solution is smooth...) break down when the interval for t contains
0. Therefore to make things simple we either work in t > 0 or in t < 0, to avoid containing t = 0.