Review of methods for finding
the particular solution of a nonhomogeneous system
Everything in this review is based on the following:
x0 = Px + g
This is also written
¸ ·
¸·
¸
¸ ·
p11 p12
x01
x1
g1 (t)
=
+
p21 p22
x02
x2
g2 (t)
| {z } |
{z
} | {z } | {z }
·
x0
P(t)
x
g(t)
When discussing the general solution to this system, I will use the following notation:
• Eigenvalues: r1 , r2 , etc.
• Eigenvectors:
· 1
¸
· 2
¸
ξ1 (t)
ξ1 (t)
1
2
ξ =
for r1 , and ξ =
for r2 .
ξ21 (t)
ξ22 (t)
• General solution:
xg (t) = c1 ξ 1 er1 t + c2 ξ 2 er2 t
·
¸
·
¸
ξ11
ξ12
r1 t
= c1
e + c2
er2 t
ξ21
ξ22
Recall: ξ12 = ξ12 , ξ21 = ξ21 , etc.
• Particular solution: xp (t).
I will try to keep the notation consistent throughout, although I may not include the (t)
every time.
I am assuming that you have already found the general solution of the system, in everything
that follows.
1. Diagonalization
Summary: The principle of diagonalization is to exploit the magic of linear algebra to
make your computations simpler. Specifically, if P is diagonalizable, let x = Ty so that you
can solve the simpler system
y0 = Dy + T−1 g.
(1)
Advantages:
a. This system is simpler because D is a diagonal matrix.
b. D does not need to be calculated. Although you can find it from computing T−1 PT,
this computation is unnecessary. D is just the matrix of eigenvalues, which you have
already found:
·
¸
r1 0
D=
.
0 r2
c. T has also already been found: it is just the matrix of eigenvalues. Note: T is not
quite the same as the fundamental matrix Ψ because it does not include the functions
er1 t , er2 t , etc.
Disadvantages:
a. The matrix P must be diagonalizable. This means that it must have n linearly
independent eigenvectors if P is n × n. Remember: P is automatically diagonalizable
if it is Hermitian, that is:
T
P=P .
If P contains only real numbers, then this just means that P is automatically diagonalizable whenever P is symmetric:
P = PT .
How to use the method of diagonalization:
1. Write down the matrix of eigenvectors T and find its inverse:
·
¸
·
¸
ξ11 ξ12
ξ22 −ξ12
−1
1
T=
T = ξ11 ·ξ22 −ξ12 ·ξ21
.
ξ21 ξ22
−ξ21
ξ11
2. Compute
·
−1
T g=
1
ξ11 ·ξ22 −ξ12 ·ξ21
ξ22 −ξ12
−ξ21
ξ11
¸·
g1 (t)
g2 (t)
¸
·
=
h1 (t)
h2 (t)
¸
.
3. Optional step: use the principle of diagonalization (1). Write out y0 = Dy + T−1 g :
¸
· 0 ¸ ·
r1 y1 + h1 (t)
y1
=
r2 y2 + h2 (t)
y20
This step is not strictly necessary, but sometimes it’s good to keep the formalism in
mind as a way of remembering where you are and what you’re doing.
Also, it shows the TA that you know what’s up.
4. Find y1 , y2 by evaluating the indefinite integrals:
Z
r1 t
y1 = e
e−r1 s h1 (s) ds
Z
r2 t
y2 = e
e−r2 s h2 (s) ds
Tip: don’t need to bother keeping track of the constants of integration. They disappear at the end anyway and they just make more work in the mean time.
5. Now that you have y, you can find the particular solution to the original problem by
computing
·
¸·
¸
ξ11 ξ12
y1
xp = Ty =
.
ξ21 ξ22
y2
Common mistakes to avoid:
1. In step 4, integrate only the e−r1 s hj (t). Don’t include the rj yj part.
2. In step 4, don’t forget to multiply the integral by er1 t .
3. DON’T forget step 5. For some reason, people really like to forget step 5.
2. Undetermined Coefficients
Summary: The principle of undetermined coefficients is to exploit the tractability of the
functions in g(t) to make your computations simpler. Specifically, we assume that a particular solution must look like
(2)
xp = aγ1 (t) + bγ2 (t) + dγ3 (t) + e.
By comparing the coefficients of the γ1 (t) term we can figure out what a is, etc.
Advantages:
a. Easy mathematics: involves NO integration.
b. Can be used to find a particular solution without having to find eigenvectors.
Disadvantages:
a. Only works when g contains nice functions (polynomials, exponentials, trigs).
b. P must have constant coefficients.
How to use the method of undetermined coefficients:
1. Write out
·
¸ ·
¸
·
¸
·
¸
g1
α1
α2
α3
g(t) =
=
γ1 (t) +
γ2 (t) +
γ3 (t).
g2
β1
β2
β3
Here, αj , βj are numbers and γj are functions of t like ekt , cos kt, sin kt, t, t2 , etc. I’ve
written g as if it had three different terms, but of course it probably only has one or
two different terms (and could theoretically have many more).
2. Use the principle of undetermined coefficients (2) and write out
xp = aγ1 (t) + bγ2 (t) + dγ3 (t) + e.
Note: if an eigenvalue r = ri shows up in a γj (t) = ert , you must assume x includes
an extra term with a t, e.g.,
xp = aert + btert + dγ2 (t) + eγ3 (t) + f .
3. Find the derivative
x0p = aγ10 (t) + bγ20 (t) + dγ30 (t).
4. Rewrite x0 = Px + g as
aγ10
|
+
bγ20
{z
x0p
+
dγ30
}
·
= Paγ1 + Pbγ2 + Pdγ3 + Pe +
{z
}
|
Pxp
|
α1
β1
¸
·
γ1 +
¸
·
¸
α2
α3
γ2 +
γ3 .
β2
β3
{z
}
g
5. Collect all the γj terms to obtain equations like
·
¸
·
¸
α1
α1
0
γ1
=⇒
Pa = a −
.
Paγ1 = aγ1 −
β1
β1
Note: these equations may look quite different from what I’ve written here, e.g., Pa
may not always be with aγ10 . The important thing is to gather all the terms with
γ1 together, all the terms with γ2 together, etc. Then you can make an equation
between the coefficients of the γj . (This is where the name of this method comes
from, right?)
6. Solve these equations to find a1 , a2 :
·
¸·
¸ ·
¸ ·
¸
p11 p12
a1
p11 a1 + p12 a2
a1 − α1
=
=
p21 p22
a2
p21 a1 + p22 a2
a2 − β1
·
=⇒
a=
a1
a2
¸
Sometimes, you may not be able to solve for a completely. For example, suppose you
obtain
·
¸
a1
a=
a1 − 15
from the equations you obtained in part 5. In this case, follow the book and write
· ¸ ·
·
¸
¸
0
a1
1
+
a=
= a1
.
1
a1 − 15
− 15
Then set a1 = 0 and take
·
a=
0
¸
.
− 15
Also note that sometimes you may need to solve for b before you can solve for a, etc.
7. Now that you have found a, b,etc., write out the particular solution
·
¸
·
¸
·
¸
·
¸
a1
b1
d1
e1
xp (t) =
γ1 (t) +
γ2 (t) +
γ3 (t) +
.
a2
b2
d2
e2
Note: this is the particular solution, not the general solution. THERE ARE NO
CONSTANTS c1 , c2 in front of these terms.
Common mistakes to avoid:
1. DON’T forget the g in step 4. People love to forget the g in step 4.
2. The g starts off on the same side as the Px is in step 4. It needs to be subtracted
over to the side in step 5. Many people make a sign error at this point.
3. Variation of Parameters
Summary: exploit the properties of the fundamental matrix Ψ by solving the equation
(3)
Ψu0 = g.
to obtain a vector u. Then the particular solution is xp = Ψu.
Advantages:
a. May be applied to solve any problem.
b. Easy to remember: the only formulae you need are are Ψu0 = g and xp = Ψu.
Disadvantages:
a. Solving Ψu0 = g can get ugly because each component of the vectors involved can
contain a complicated expression and row-reduction becomes cumbersome.
b. Integrating u0 to find u can sometimes be difficult.
c. Doing the matrix multiplication Ψu can be tedious.
How to use the method of variation of parameters:
1. Use your general solution to form the fundamental matrix, which probably looks like
this:
·
¸
ξ11 er1 t ξ12 er2 t
Ψ=
.
ξ21 er1 t ξ22 er2 t
2. Use the principle of variation of parameters (3) and solve the matrix equation Ψu0 = g
to find u0 :
·
¸· 0
¸ ·
¸
ξ11 er1 t ξ12 er2 t
u1 (t)
g1 (t)
=
.
ξ21 er1 t ξ22 er2 t
u02 (t)
g2 (t)
3. Integrate u01 to find u1 , and integrate u02 to find u2 .
Tip: You don’t need to worry about the constants of integration here, as they will
magically disappear in the last step anyway.
4. Now that you have u, perform the matrix multiplication Ψu to recover xp :
·
¸·
¸ ·
¸
ξ11 er1 t ξ12 er2 t
u1 (t)
x1 (t)
=
= xp (t).
ξ21 er1 t ξ22 er2 t
u2 (t)
x2 (t)
Common mistakes to avoid:
1. Don’t forget step 4. People love to forget step 4.