Download Solving linear, const.-coeff. ODEs

Solving linear, const.-coeff. ODEs
This is a summary of the method for finding
the general solution of linear, first and second
order, constant-coefficient ODEs. Details of the
theory can be found in most introductory books
on ordinary differential equations such as Boyce
& DiPrima.
The ODE to be solved for y = y(x) is of the
form
a y 00 + b y 0 + c y = F (x) ,
(1)
where a , b and c are real constants (thus the
classification as a constant-coefficient equation),
and F (x) is a given (forcing) function.
Note: If F (x) ≡ 0 the equation is homogeneous, otherwise it is non-homogeneous.
1
First order equations
If a = 0 and b 6= 0 , then the equation is a first
order ODE
b y 0 + c y = F (x) ,
(2)
and its general solution is
1 x −c(x−ξ)/b
−c(x−x
)/b
0
y(x) = y(x0) e
+
e
F (ξ) dξ ,
b x0
where x0 can be chosen arbitrarily.
Z
This is obtained multiplying (2) by the integrating factor
1 cx/b
e
.
b
It can also be obtained by the method described next for
µ(x) =
2nd order equations.
Homogeneous 2nd order equations
If a 6= 0 the ODE is 2nd order, and the general solution of (1) can be decomposed into the
sum of the general solution of the corresponding
homogeneous equation,
a y 00 + b y 0 + c y = 0 ,
(3)
and one particular solution of the non-homogeneous
equation (1). Thus we have
y(x) = yh(x) + yp(x) ,
where yh(x) is the general solution of (3), and
yp(x) is a solution of (1). In this section a
method to find yh is presented, and in subsequent sections methods to find yp will be discussed — they can also be found in most introductory books. Variation of parameters, and
the method of undetermined coefficients will be
reviewed in this document.
Assume that the solution of (3) has the form
y(x) = erx and plug into (3) to conclude that r
must be a root of the characteristic equation
a r2 + b r + c = 0 .
(4)
The form of yh depends on the roots of (4).
The 3 possible cases will be discussed next.
Real, distinct roots
Suppose b2 − 4ac > 0 . In this case there are two
distinct real roots of the characteristic equation,
r
2
b − 4ac
b
r1,2 = −
±
,
2a
2a
and the general solution of the homogeneous
equation (3) is
yh(x) = A1 er1x + A2 er2x ,
where A1 and A2 are arbitrary constants which
are usually determined from initial and/or boundary conditions. An alternative form of the general solution of (3) which is sometimes preferred
is
yh(x) = eλx {B1 cosh(µx) + B2 sinh(µx)} , (5)
where B1 and B2 are arbitrary constants, and
b
λ=− ,
2a
µ=
r
2
b − 4ac
2a
,
r1,2 = λ ± µ .
(6)
As an example, consider the following ODE
y 00 − y 0 − 2y = 0
which has characteristic equation
r2 −r−2 = 0 ,
with roots
r1 = 2 ,
r2 = −1 .
The general solution of the ODE can be written
as
y(x) = A1 e2x + A2 e−x ,
or as (here λ = 1/2 , µ = 3/2 and r1,2 = λ ± µ )
y(x) = ex/2 {B1 cosh(3x/2) + B2 sinh(3x/2)} .
Real, equal roots
If b2 − 4ac = 0 , then the roots of the characteristic equation are
b
,
2a
and the general solution of (3) is
r1 = r2 = λ = −
yh(x) = eλx {A1 x + A2} .
As an example consider the ODE
y 00 + 4y 0 + 4y = 0 .
The characteristic equation r2 + 4r + 4 = 0 has
only one root: r1 = r2 = λ = −2 , and the
general solution of the ODE is
y(x) = e−2x {A1 x + A2} .
Complex conjugate roots
If b2 − 4ac < 0 , then the roots of the characteristic equation are complex conjugate numbers
r1,2 = λ ± iµ ,
where λ and µ are given in (6). In this case, the
general solution of the homogeneous equation
(3) is (note the similarity with (5))
yh(x) = eλx {A1 cos(µx) + A2 sin(µx)} .
As an example consider the ODE
y 00 + 4y 0 + 13y = 0 .
The roots of the char. equa. are r1,2 = −2 ± 3i
(here λ = −2 , µ = 3), and the general solution
is
y(x) = e−2x {A1 cos(3x) + A2 sin(3x)} .
Constant-coefficient case: Method of undetermined coefficients
Let L[y] := ay 00 + by 0 + cy . The goal is to find a
particular solution yp(x) to the non-homogeneous
equation (1)
L[y] = F (x) ,
ay 00 + by 0 + cy = F (x) .
i.e.
In order to do this, note that if the forcing function is a sum
F (x) =
m
X
Fj (x) ,
(7)
j=1
then we can decompose yp(x) as a sum
yp(x) =
m
X
ypj (x) ,
(8)
j=1
where
L[ypj ] = Fj (x) ,
j = 1, . . . , m .
(9)
Indeed, from (8),(9),(7) and the linearity of the
ODE we obtain

L[yp] = L 
m
X
j=1

ypj  =
m
X
j=1
L[ypj ] =
m
X
j=1
Fj (x) = F (x) .
The method of undetermined coefficients works
for the case of constant-coefficient, linear ODEs,
when the forcing function is of a special form as
described below. Assume that
or
F (x) = Pn(x) eα x cos(β x) ,
(10)
F (x) = Pn(x) eα x sin(β x) ,
(11)
where Pn(x) is a polynomial of degree n , and α ,
β are real numbers. Assume that α+iβ is a root
of multiplicity s of the characteristic equation
ar 2 + br + c = 0 , i.e.
• s = 0 means that α + iβ is not a root of the
characteristic equation,
• if s = 1 then α + iβ is a simple root of the
characteristic equation,
• if s = 2 then α + iβ is a double root of the
characteristic equation.
In the cases (10),(11), the form of the particular
solution is
yp(x) = xs eα x [Qn(x) cos(β x) + Rn(x) sin(β x)] ,
where Qn(x) and Rn(x) are polynomials of degree n (same degree as Pn(x)) with coefficients
to be determined from (1), i.e.
Qn(x) = An xn + An−1 xn−1 + · · · + A1 x + A0 ,
Rn(x) = Bn xn + Bn−1 xn−1 + · · · + B1 x + B0 ,
where An, . . . , A0 and Bn, . . . , B0 have to be determined from (1). Note some special cases:
(a) If α = β = 0, then s is the multiplicity of 0
(zero) as a root of the characteristic equation;
(b) if β = 0, then s is the multiplicity of α as a
root of the characteristic equation; (c) if α = 0,
then s is the multiplicity of iβ as a root of the
characteristic equation. The following examples
will illustrate the method.
Example 1
L[y] := y 00 + 5y 0 + 6y = 4ex cos(2x)
(12)
The char. equa. has roots r1 = −2 , r2 = −3.
Therefore, 1+2i is not a root of the char. equa.,
and s = 0. So, we should try (here α = 1, β = 2,
n = 0 in (10))
yp(x) = ex [A0 cos(2x) + B0 sin(2x)] .
Then,
yp0 (x) = ex [(A0 + 2B0) cos(2x) + (B0 − 2A0) sin(2x)]
yp00(x) = ex [(4B0 − 3A0) cos(2x)
−(4A0 + 3B0) sin(2x)] ,
so that
L[yp] = ex [(8A0 + 14B0) cos(2x)
− (14A0 − 8B0) sin(2x)] .
Thus, yp satisfies (12) provided
8A0 + 14B0 = 4
14A0 − 8B0 = 0
i.e. A0 = 8/65 and B0 = 14/65 , and
ex
yp(x) =
[8 cos(2x) + 14 sin(2x)] .
65
(13)
Example 2
L[y] := y 00 + 5y 0 + 6y = −5x sin(3x)
(14)
The char. equa. has roots r1 = −2 , r2 = −3,
and 3i is not a root of the char. equa., so s = 0
and we should try (here α = 0, β = 3, n = 1)
yp(x) = (A1x+A0) cos(3x)+(B1x+B0) sin(3x) .
Then,
yp0 (x) = (A1 + 3B1x + 3B0) cos(3x)
+ (B1 − 3A1x − 3A0) sin(3x)
yp00(x) = (6B1 − 9A1x − 9A0) cos(3x)
− (6A1 + 9B1x + 9B0) sin(3x) ,
so that
L[yp] = [(15B1 − 3A1)x + (5A1 + 6B1 − 3A0
+15B0)] cos(3x)
− [(15A1 + 3B1)x + (6A1 − 5B1
+15A0 + 3B0)] sin(3x) .
Thus, yp satisfies (14) provided
15B1 − 3A1
5A1 + 6B1 − 3A0 + 15B0
15A1 + 3B1
6A1 − 5B1 + 15A0 + 3B0
=0
=0
=5
=0
i.e. A0 = −235/3042 , B0 = −25/169 A1 = 25/78 ,
B1 = 5/78 , and
235
25
x−
cos(3x)
yp(x) =
78
3042
5
25
x−
sin(3x) . (15)
+
78
169
Example 3
L[y] := y 00 + 5y 0 + 6y = 2x2e−3x
(16)
The char. equa. has roots r1 = −2 , r2 = −3,
and −3 is a simple root of the char. equa., so
s = 1 and we should try (α = −3, β = 0, n = 2)
yp(x) = x (A2 x2 + A1 x + A0) e−3x .
Then,
h
0
yp(x) = (−3A2 x3 + (3A2 − 3A1)x2 + (2A1 − 3A0)x
i
+ A0 e−3x
h
00
yp (x) = 9A2 x3 − (18A2 − 9A1)x2
i
+ (6A2 − 12A1 + 9A0)x + (2A1 − 6A0) e−3x
so that
h
L[yp] = − 3A2 x2 + (6A2 − 2A1)x
+ (2A1 − A0) e−3x .
i
Thus, yp satisfies (16) provided
−3A2 = 2
6A2 − 2A1 = 0
2A1 − A0 = 0
i.e. A0 = −4 , A1 = −2 A2 = −2/3 , and
2 2
yp(x) = −x
x + 2 x + 4 e−3x .
3
Example 4
(17)
L[y] := y 00 + 5y 0 + 6y = 2x2e−3x − 5x sin(3x)
+ 4ex cos(2x)
Since the forcing function here is a sum of the
forcing functions in (12),(14) and (16), we conclude from (8),(13),(15) and (17) that
2 2
x + 2 x + 4 e−3x
yp(x) = −x
3
25
235
+
x−
cos(3x)
78
3042
5
25
+
x−
sin(3x)
78
169
ex
+
[8 cos(2x) + 14 sin(2x)] .
65
General case: Method of variation of parameters
This method relies on the knowledge of a fundamental set of solutions of the homogeneous
linear equation
y 00 + p(t) y 0 + q(t) y = 0 ,
say { y1(t), y2(t) }, to obtain a particular solution of the non-homogeneous equation (in normal form)
y 00 + p(t) y 0 + q(t) y = F (t) .
(18)
The idea is to postulate a form of the particular
solution as
yp(t) = u(t)y1(t) + v(t)y2(t) ,
(19)
and to find the (variable) parameters u and v
so that yp(t) satisfies (18). The ODE gives one
condition on the two parameters. We are free
to impose an additional condition for these two
unknowns. The following system of equations is
obtained.
y1 u0 + y2 v 0 = 0
y10 u0 + y20 v 0 = F (t)
(20)
(21)
Since the coefficient matrix of this algebraic system has determinant equal to the Wronskian of
the fundamental set (which is nonzero)
y
1 y2 0 − y0 y ,
=
y
y
W (t) := W [y1, y2] = 0
1
0
2
1 2
y1 y2
there is a unique solution for u0 and v 0 , which
can be expressed (Cramer’s rule) as
u0(t) =
v 0(t) =
0
y2
0 F (t) y2
W (t)
y
0 1
0
y1 F (t)
W (t)
−y2(t)F (t)
=
,
W (t)
y1(t)F (t)
=
.
W (t)
Hence,
u(t) = −
v(t) =
Z
Z
y2(t)F (t)
dt ,
W (t)
y1(t)F (t)
dt .
W (t)
(22)
(23)
Example 5
t2y 00 − 3ty 0 + 4y = t2 ln t ,
t > 0.
(24)
It is easy to verify that y1 = t2 and y2 = t2 ln t
form a fundamental set of solutions of the corresponding homogeneous ODE on the interval
(0, ∞). Also, the normal form of (24) is
3 0
4
00
y − y + 2 y = ln t ,
t
t
t > 0.
Therefore, the method of variation of parameters gives — see (19)
yp = u y 1 + v y 2 ,
where u0 and v 0 satisfy the system of algebraic
equations (20),(21)
t2 u0 + t2(ln t) v 0 = 0 ,
2t u0 + t(1 + 2 ln t) v 0 = ln t ,
which has the solution (22),(23) (you can solve
the system above any way you want)
(ln t)2
0
u =−
,
t
Upon integration, we obtain
(ln t)3
u(t) = −
,
3
v0 =
ln t
.
t
(ln t)2
v(t) =
,
2
so that
t2(ln t)3
t2(ln t)3
+
yp(t) = −
3
2
t2(ln t)3
or
yp(t) =
.
6
Thus, the general solution of (24) is
t2 (ln t)3
2
2
y = c1 t + c2 t ln t +
,
6
or
y = t2
c1 + c2 ln t +
(ln t)3
6
!
.