Download Characteristic Equation & Factoring of Linear Homogeneous Diff. Equations and more Summaries Algebra in PDF only on Docsity!
Arbitrary Homogeneous Linear
Equations with Constant Coefficients
In chapter 16, we saw how to solve any equation of the form
ay′′^ + by′^ + cy = 0
when a , b and c are real constants. Unsurprisingly, the same basic ideas apply when dealing with any equation of the form
a 0 y(N^ )^ + a 1 y(N^ −^1 )^ + · · · + aN − 2 y′′^ + aN − 1 y′^ + aN y = 0
when N is some positive integer and the ak ’s are all real constants. Assuming y = er x^ still leads to the corresponding “characteristic equation”
a 0 r N^ + a 1 r N^ −^1 + · · · + aN − 2 r 2 + aN − 1 r + aN = 0 ,
and a general solution to the differential equation can then be obtained using the solutions to the characteristic equation, much as we did in chapter 16. Computationally, the only significant difficulty is in the algebra needed to find the roots of the characteristic polynomial. So let us look at that algebra, first.
18.1 Some Algebra
A basic fact of algebra is that any second-degree polynomial
p(r ) = ar 2 + br + c
can be factored to p(r ) = a(r − r 1 )(r − r 2 )
where r 1 and r 2 are the roots of the polynomial (i.e., the solutions to p(r ) = 0 ). These roots may be complex, in which case r 1 and r 2 are complex conjugates of each other (assuming a , b and c are real numbers). It is also possible that r 1 = r 2 , in which case the factored form of the polynomial is more concisely written as
p(r ) = a(r − r 1 )^2.
375
376 Arbitrary Homogeneous Linear Equations with Constant Coefficients
The idea of “factoring”, of course, extends to polynomials of higher degree. And to use this idea with these polynomials, it will help to introduce the “completely factored form” for an arbitrary N th-degree polynomial
p(r ) = a 0 r N^ + a 1 r N^ −^1 + · · · + aN − 2 r 2 + aN − 1 r + aN.
We will say that we’ve (re)written this polynomial into its completely factored form if and only if we’ve factored it to an expression of the form
p(r ) = a 0 (r − r 1 )m^1 (r − r 2 )m^2 · · · (r − rK )m^ K^ (18.1)
where { r 1 , r 2 ,... , rK }
is the set of all different (possibly complex) roots of the polynomial (i.e., values of r satisfying p(r ) = 0 ), and { m 1 , m 2 ,... , m (^) K }
is some corresponding set of positive integers. Let’s make a few simple observations regarding the above, and then look at a few exam- ples:
- It will be important for our discussion that
{ r 1 , r 2 ,... , rK }
is the set of all different roots of the polynomial. If j 6 = k , then r (^) j 6 = rk.
- Each mk is the largest integer such that (r − rk )mk^ is a factor of the original polynomial. Consequently, for each rk , there is only one possible value for mk. We call mk the multiplicity of rk.
- As shorthand, we often say that rk is a simple root if its multiplicity is 1 , a double root if its multiplicity is 2 , a triple root if its multiplicity is 3 , and so on.
- If you multiply out all the factors in the completely factored form in line (18.1), you get a polynomial of degree m 1 + m 2 + · · · + m (^) K. Since this polynomial is supposed to be p(r ) , an N th-degree polynomial, we must have
m 1 + m 2 + · · · + m (^) K = N.
! ◮ Example 18.1: By straightforward multiplication, you can verify that
2 (r − 4 )^3 (r + 5 ) = 2 r 4 − 14 r 3 − 24 r 2 + 352 r − 640.
This means p(r ) = 2 r 4 − 14 r 3 − 24 r 2 + 352 r − 640 can be written in completely factored form
p(r ) = 2 (r − 4 )^3 (r − [− 5 ]).
This polynomial has two distinct real roots, 4 and − 5. The root 4 has multiplicity 3 , and − 5 is a simple root.
378 Arbitrary Homogeneous Linear Equations with Constant Coefficients
18.2 Solving the Differential Equation
The Characteristic Equation
Suppose we have some N th-order differential equation of the form
a 0 y(N^ )^ + a 1 y(N^ −^1 )^ + · · · + aN − 2 y′′^ + aN − 1 y′^ + aN y = 0 (18.2)
where the ak ’s are all constants (and a 0 6 = 0 ). Since ( er x^
= r er x ( er x^
r er x^
= r · r er x^ = r 2 er x ( er x^
r 2 er x^
= r 2 · r er x^ = r 3 er x .. .
for any constant r , it is easy to see that plugging y = er x^ into the differential equation yields
a 0 r N^ er x^ + a 1 r N^ −^1 er x^ + · · · + aN − 2 r 2 er x^ + aN − 1 r er x^ + aN er x^ = 0 ,
which, after dividing out er x^ , gives us the corresponding characteristic equation
a 0 r N^ + a 1 r N^ −^1 + · · · + aN − 2 r 2 + aN − 1 r + aN = 0. (18.3)
As before, we refer to the polynomial on the left,
p(r ) = a 0 r N^ + a 1 r N^ −^1 + · · · + aN − 2 r 2 + aN − 1 r + aN ,
as the characteristic polynomial for the differential equation. Also, as in a previous chapter, it should be observed that the characteristic equation can be obtained from the original differential equation by simply replacing the derivatives of y with the corresponding powers of r. According to the complete factorization theorem, the above characteristic equation can be rewritten in completely factored form,
a 0 (r − r 1 )m^1 (r − r 2 )m^2 · · · (r − rK )m^ K^ = 0 (18.4)
where the rk ’s are all the different roots of the characteristic polynomial, and the mk ’s are the multiplicities of the corresponding roots. It turns out that, for each root rk with multiplicity mk , we can identify a corresponding linearly independent set of mk particular solutions to the original differential equation. It will be obvious (once you see them) that no solution generated from one root can be written as a linear combination of solutions generated from the other roots. Hence, the set of all these particular solutions generated from all the rk ’s will be a linearly independent set containing (according to our complete factorization theorem)
m 1 + m 2 + · · · + m (^) K = N
solutions. From the big theorem on solutions to homogeneous equations (theorem 14.2 on page 304), we then know that this big set is a fundamental set of solutions for the differential equation, and that the general solution is given by an arbitrary linear combination of these particular solutions. Exactly which particular solutions are generated from each individual root depends on the multiplicity and whether the root is real valued or not.
Solving the Differential Equation 379
Particular Solutions Corresponding to One Root
In the following, we will assume rk is a root of multiplicity mk to our characteristic polynomial. That is, (r − rk )mk
is one factor in equation (18.4). However, since the choice of k will be irrelevant in this discussion, we will, for simplicity, drop the subscripts.
The Basic Result
Assume r is a root of multiplicity m to our characteristic polynomial. Then, as before,
er x
is one particular solution to the differential equation, and if m = 1 , it is the only solution corresponding to this root we need to find. So now assume m > 1. In the previous chapter, we found that
x er x
is a second solution to the differential equation when r is a repeated root and N = 2. This was obtained via reduction of order. For the more general case being considered here, it can be shown that x er x^ is still a solution. In fact, it can be shown that the m particular solutions to the differential equation corresponding to root r can be generated one after the other by simply multiplying the previously found solution by x. That is, we have the following theorem:
Theorem 18. Let r be a root of multiplicity m to the characteristic polynomial for
a 0 y(N^ )^ + a 1 y(N^ −^1 )^ + · · · + aN − 2 y′′^ + aN − 1 y′^ + aN y = 0
where the ak ’s are all constants. Then { er x^ , x er x^ , x^2 er x^ ,... , xm−^1 er x^
is a linearly independent set of m solutions to the differential equation.
The proof of this theorem will be discussed later, in section 18.4. (And it probably should be noted that xm^ er x^ ends up not being a solution to the differential equation.)
Particular Solutions Corresponding to a Real Root
If r is a real root of multiplicity m to our characteristic polynomial, then theorem 18.2, above, tells us that (^) { er x^ , x er x^ , x^2 er x^ ,... , xm−^1 er x^
is the linearly independent set of m solutions to the differential equation corresponding to that root. No more need be said.
! ◮ Example 18.4: Consider the homogeneous differential equation
y(^5 )^ − 15 y(^4 )^ + 90 y′′′^ − 270 y′′^ + 405 y′^ − 243 y = 0.
Solving the Differential Equation 381
For many problems, though, these are not particularly desirable sets of solutions because they introduce complex values into computations we expect to yield real values. But recall how we dealt with complex-exponential solutions for second-order equations, constructing alterna- tive pairs of solutions via linear combinations. Let us try the same idea here, constructing an alternative pair of solutions {yk, 1 , yk, 2 } from each pair { xk^ e(λ+i^ ω)x^ , xk^ e(λ−i^ ω)x^
by the linear combinations
yk, 1 (x) = 12 xk^ e(λ+i^ ω)x^ + 12 xk^ e(λ−i^ ω)x
and
yk, 2 (x) = 1 2 i xk^ e(λ+i^ ω)x^ − 1 2 i xk^ e(λ−i^ ω)x^.
Since e(λ±i^ ω)x^ = eλx^ [cos(ωx) ∓ i sin(ωx)] ,
you can easily verify that
yk, 1 = xk^ eλx^ cos(ωx) and yk, 2 = xk^ eλx^ sin(ωx).
It is also “easily” verified that the set of these functions, with k = 0 , 1 , 2 ,... , m − 1 , is linearly independent. Thus, instead of using { e(λ+i^ ω)x^ , x e(λ+i^ ω)x^ , x^2 e(λ+i^ ω)x^ ,... , xm−^1 e(λ+i^ ω)x^
and { e(λ−i^ ω)x^ , x e(λ−i^ ω)x^ , x^2 e(λ−i^ ω)x^ ,... , xm−^1 e(λ−i^ ω)x^
as the two linearly independent sets corresponding to roots λ + iω and λ − iω , we can use the sets of real-valued functions { eλx^ cos(ωx) , x eλx^ cos(ωx) , x^2 eλx^ cos(ωx) ,... , xm−^1 eλx^ cos(ωx)
and { eλx^ sin(ωx) , x eλx^ sin(ωx) , x^2 eλx^ sin(ωx) ,... , xm−^1 eλx^ sin(ωx)
! ◮ Example 18.5: Consider the differential equation
y(^4 )^ − 12 y(^3 )^ + 86 y′′^ − 300 y′^ + 625 y = 0.
Its characteristic equation is
r 4 − 12 r 3 + 86 r 2 − 300 r + 625 = 0 ,
which, as we saw in example 18.3, can be factored to ( r − [ 3 + 4 i]
r − [ 3 − 4 i]
Here, we have a conjugate pair of roots, 3 + 4 i and 3 − 4 i , each with multiplicity 2. So the corresponding particular real-valued solutions to the differential equation are
e^3 x^ cos( 4 x) , xe^3 x^ cos( 4 x) , e^3 x^ sin( 4 x) and xe^3 x^ sin( 4 x).
382 Arbitrary Homogeneous Linear Equations with Constant Coefficients
And since our homogeneous, linear differential equation is of order 4 , its general solution is given by an arbitrary linear combination of these four solutions,
y(x) = c 1 e^3 x^ cos( 4 x) + c 2 xe^3 x^ cos( 4 x) + c 3 e^3 x^ sin( 4 x) + c 4 xe^3 x^ sin( 4 x) ,
which, to save space, might also be written as
y(x) = [c 1 + c 2 x]e^3 x^ cos( 4 x) + [c 3 + c 4 x]e^3 x^ sin( 4 x).
18.3 Some More Examples
The most difficult part of solving a high-order, homogeneous linear differential equation with constant coefficients is the factoring of its characteristic polynomial. Unfortunately, the methods commonly used to factor second-degree polynomials do not nicely generalize to methods for factoring polynomials of higher degree. So we have to use whatever algebraic tricks we can think of. And if all else fails, we can run to the computer and let our favorite math package attempt the factoring. Here are a few examples to help you recall some of the useful tricks for factoring polynomials of order three or above.
! ◮ Example 18.6: Consider the seventh-order, homogeneous differential equation
y(^7 )^ − 625 y(^3 )^ = 0.
The characteristic equation is r 7 − 625 r 3 = 0. An obvious choice of action would be to first factor out r 3 ,
r 3
r 4 − 625
Cleverly noting that r 4 = [r 2 ]^2 and 625 = 252 , and then applying well-known algebraic formulas, we have
r 3
r 4 − 625
֒ → r 3
[r 2 ]^2 − [ 25 ]^2
֒ → r 3
r 2 − 25
r 2 + 25
֒ → r 3 (r − 5 )(r + 5 )
r 2 + 25
Now
r 2 + 25 = 0 H⇒ r 2 = − 25 H⇒ r 2 = ±
− 25 = ± 5 i.
384 Arbitrary Homogeneous Linear Equations with Constant Coefficients
Trying r 1 = 1 , we get
p( 1 ) = 13 − 19 · 1 + 30 = 12 6 = 0.
So r 1 6 = 1. Trying r 1 = − 1 , we get
p(− 1 ) = (− 1 )^3 − 19 · (− 1 ) + 30 = − 1 + 19 + 30 6 = 0.
So r 1 6 = − 1. Trying r 1 = 2 , we get
p( 2 ) = ( 2 )^3 − 19 · ( 2 ) + 30 = 8 − 38 + 30 = 0.
Success! One root is r 1 = 2 and one factor of our characteristic polynomial is (r − 2 ). To get our first factoring, we then divide r − 2 into the characteristic polynomial:
r 2 + 2 r − 15 r − 2
r 3 − 19 r + 30 − r 3 + 2 r 2 2 r 2 − 19 r − 2 r 2 + 4 r − 15 r + 30 15 r − 30 0
Thus, r 3 − 19 r + 30 = (r − 2 )(r 2 + 2 r − 15 ). By inspection, we see that
r 2 + 2 r − 15 = (r + 5 )(r − 3 ).
So, our characteristic equation
r 3 − 19 r + 30 = 0
factors to (r − 2 )(r − [− 5 ])(r − 3 ) = 0 , and, thus, y(x) = c 1 e^2 x^ + c 2 e−^5 x^ + c 3 e^3 x is a general solution to our differential equation.
On Verifying Theorem 18.2 385
18.4 On Verifying Theorem 18.
Theorem 18.2 claims to give a linearly independent set of solutions to a linear homogeneous differential equation with constant coefficients corresponding to a repeated root for the equation’s characteristic polynomial. Our task of verifying this claim will be greatly simplified if we slightly expand our discussion of “factoring” linear differential operators from section 12.4. (You may want to go back and quickly review that section.)
Linear Differential Operators with Constant Coefficients
First, we need to expand our terminology a little: When we refer to L as being an N th-order linear differential operator with constant coefficients, we just mean that L is an N th-order linear differential operator
L = a 0 d N dx N^
- a 1 d N^ −^1 dx N^ −^1
- · · · + aN − 2 d^2 dx^2
- aN − 1 d dx
- aN
in which all the ak ’s are constants. Its characteristic polynomial p(r ) is simply the polynomial
p(r ) = a 0 r N^ + a 1 r N^ −^1 + · · · + aN − 2 r 2 + aN − 1 r + aN.
It turns out that factoring a linear differential operator with constant coefficients is remarkably easy if you already have the factorization for its characteristic polynomial.
! ◮ Example 18.8: Consider the linear differential operator
L = d^2 dx^2
d dx
It’s characteristic polynomial is r 2 − 5 r + 6 , which factors to (r − 2 )(r − 3 ). Now, consider the analogous composition product ( d dx
d dx
Letting φ be any suitably differentiable function, we see that ( d dx
d dx
[φ] =
d dx
) [
dφ dx − 3 φ
]
d dx
[
dφ dx − 3 φ
]
[
dφ dx − 3 φ
]
d^2 φ dx^2
dφ dx
dφ dx
d^2 φ dx^2
dφ dx
= L[φ].
On Verifying Theorem 18.2 387
there is a corresponding factorization of that operator. To simplify writing the factors of the operator when rk is a multiple root of the characteristic polynomial, let us agree that
( d dx − r 1
d dx − r 1
d dx − r 1
d dx − r 1
d dx − r 1
d dx − r 1
d dx − r 1
d dx − r 1
d dx − r 1
d dx − r 1
Using this notation along with the obvious extension of the above proof yields the next theorem.
Theorem 18.5 (factorization of constant coefficient operators) Let L be a linear differential operator with constant coefficients, and let
p(r ) = a(r − r 1 )m^1 (r − r 2 )m^2 · · · (r − rK )m^ K
be the completely factored form for the characteristic polynomial for L. Then
L = a
d dx − r 1
)m 1 ( d dx − r 2
)m 2 · · ·
d dx − rK
)m (^) K .
Let us make two observations regarding the polynomial p , one of the roots r (^) j of this polynomial, and the operator L from the last theorem:
- Because the order in which we write the factors of a polynomial is irrelevant, we have
p(r ) = a(r − r 1 )m^1 (r − r 2 )m^2 · · · (r − rK )m^ K^ = ap (^) j (r )(r − r (^) j )m^ j
where p (^) j (r ) is the product of all the (r − rk )’s with rk 6 = r (^) j. Hence, L can be factored by L = aL (^) j
d dx − r (^) j
)m (^) j
where L (^) j is the composition product of all the
(d /dx − rk
’s with rk 6 = r (^) j.
- If y is a solution to (^) ( d dx − r (^) j
)m (^) j [y] = 0 ,
then
L[y] = aL (^) j
d dx − r (^) j
)m (^) j [y] = aL (^) j
[(
d dx − r (^) j
)m (^) j [y]
]
= aL (^) j [ 0 ] = 0.
388 Arbitrary Homogeneous Linear Equations with Constant Coefficients
Together, these observations give us the actual result we will use.
Corollary 18. Let L be a linear differential operator with constant coefficients; let
p(r ) = a(r − r 1 )m^1 (r − r 2 )m^2 · · · (r − rK )m^ K
be the completely factored form for the characteristic polynomial for L , and let r (^) j be any one of the roots of p. Suppose, further, that y is a solution to
( d dx − r (^) j
)m (^) j [y] = 0.
Then y is also solution to L[y] = 0.
Proof of Theorem 18.
Theorem 18.2 claims that, if r is a root of multiplicity m for the characteristic polynomial of some linear homogeneous differential equation with constant coefficients, then
{ er x^ , x er x^ , x^2 er x^ ,... , xm−^1 er x^
is a linearly independent set of solutions to that differential equation. First we will verify that each of these xk^ er x^ ’s is a solution to the differential equation. Then we will confirm the linear independence of this set.
Verifying the Solutions
If you look back at corollary 18.6, you will see that we need only show that ( d dx − r
)m [xk^ er x^ ] = 0 (18.5)
whenever k is a nonnegative integer less than m. To expedite our main computations, we’ll do two preliminary computations. And, since at least one may be useful in a later chapter, we’ll describe the results in an easily referenced lemma.
Lemma 18. Let r , α and β be constants with α being a positive integer and β being real valued. Then
( d dx − r
)α [ er x^
]
= 0 and
d dx − r
)α [ xβ^ er x^
]
= β
d dx − r
)α− 1 [ xβ−^1 er x^
]
PROOF: For the first: ( d dx − r
)α (^) [ er x^
]
d dx − r
)α− 1 ( d dx − r
) [
er x^
]
390 Arbitrary Homogeneous Linear Equations with Constant Coefficients
which we can rewrite as
0 = c 0 er x^ + c 1 x er x^ + c 2 x^2 er x^ + · · · + cm xm−^1 er x
where the ck ’s are constants with cκ = − 1.
Dividing out er x^ reduces the above to
0 = c 0 + c 1 x + c 2 x^2 + · · · + cm− 1 xm−^1. (18.6)
Since this is supposed to hold for all x , it should hold for x = 0 , giving us
0 = c 0 + c 1 · 0 + c 2 · 02 + · · · + cm− 1 · 0 m−^1 = c 0.
Now differentiate both sides of equation (18.6) and plug in x = 0 :
d dx
[ 0 ] =
d dx
[
c 0 + c 1 x + c 2 x^2 + · · · + cm− 1 xm−^1
]
֒ → 0 = 0 + c 1 + 2 c 2 x + · · · + (m − 1 )cm− 1 xm−^2
֒ → 0 = 0 + c 1 + 2 c 2 · 0 + · · · + (m − 1 )cm− 1 · 0 m−^2
֒ → 0 = c 1.
Differentiating both sides of equation (18.6) twice and plugging in x = 0 :
d^2 dx^2
[ 0 ] =
d^2 dx^2
[
c 0 + c 1 x + c 2 x^2 + c 3 x^3 · · · + cm− 1 xm−^1
]
d dx
[
0 + c 1 + 2 c 2 x + 3 c 3 x^2 · · · + (m − 1 )cm− 1 xm−^2
]
֒ → 0 = 0 + 0 + 2 c 2 + 6 c 3 x · · · + (m − 1 )cm− 1 (m − 2 )xm−^2
֒ → 0 = 0 + 0 + 2 c 2 + 6 c 3 · 0 + · · · + (m − 1 )(m − 2 )cm− 1 · 0 m−^2
֒ → 0 = c 2.
Clearly, we can differentiate equation (18.6) again and again, plug in x = 0 , and, eventually, obtain 0 = ck for k = 0 , 1 , 2 ,... , m − 1.
But, one of these ck ’s is cκ which we know is −1 (assuming our set of xk^ er x^ ’s is linearly dependent). In other word, for our set of xk^ er x^ ’s to be linearly dependent, we must have
0 = cκ = − 1 ,
which is impossible. So our set of xk^ er x^ ’s cannot be linearly dependent. It must be linearly independent, just as theorem 18.2 claimed.
On Verifying Theorem 18.3 391
18.5 On Verifying Theorem 18.
Theorem 18.3 is a theorem about complex conjugation in the algebra of complex numbers. So let’s start with a brief discussion of that topic.
Algebra with Complex Conjugates
Recall that a complex number z is something that can be written as
z = x + iy
where x and y are real numbers, which we generally refer to as, respectively, the real and the imaginary parts of z. Along these lines, we say z is real if and only if z = x (i.e., y = 0 ), and we say z is imaginary if and only if z = iy (i.e., x = 0 ). The corresponding complex conjugate of z — denoted z∗^ — is z with the sign of its complex part switched,
z = x + iy H⇒ z∗^ = x + i(−y) = x − iy.
Note that (^) ( z∗
= (x − iy)∗^ = x + iy = z ,
and that z∗^ = z if z is real.
We will use these facts in a moment. We will also use formulas involving the complex conjugates of sums and products. To derive them, let
z = x + iy and c = a + ib
where x , y , a and b are all real, and compute out
(c + z)∗^ , c∗^ + z∗^ , (cz)∗^ and c∗z∗
in terms of a , b , x and y. You’ll quickly discover that
(c + z)∗^ = c∗^ + z∗^ and (cz)∗^ = c∗z∗^.
It then follows that ( z^2
= (^) (z · z)∗^ = z∗^ · z∗^ =
z∗
z^3
z^2 · z
z^2
· z∗^ =
z∗
· z∗^ =
z∗
Continuing along these lines, it is a straightforward exercise to confirm that, given any polynomial
c 0 zN^ + c 1 zN^ −^1 + · · · + cN − 1 z + cN ,
then
Additional Exercises 393
Now
(r − r 0 )
r − r 0 ∗
= (r − [λ + iω]) (r − [λ − iω]) = · · · = r 2 − 2 λr + ω^2.
So the coefficients of both the denominator and the numerator in the fraction defining p 1 (r ) are real-valued constants. If you think about how one actually computes this fraction (via, say, long division), you will realize that all the coefficients of p 1 (r ) must also be real. If m > 1 then (r − r 0 )m−^1 — but not (r − r 0 )m^ — will be a factor of p 1 (r ). Thus, r 0 will be a root of multiplicity m − 1 for p 1. Repeating the above arguments with p 1 replacing p leads to the conclusions that
- r 0 ∗^ is also a root of p 1
and
- there is an (N − 4 )th^ degree polynomial p 2 with real coefficients such that p(r ) = p 1 (r ) (r − r 0 )
r − r 0 ∗
= p 2 (r ) (r − r 0 )^2 (r − r 0 ∗)^2. Clearly, we can continue repeating these arguments, ultimately obtaining the formula p(r ) = pm (r ) (r − r 0 )m^ (r − r 0 ∗)m
where pm is a polynomial of degree N − 2 m with just real coefficients and for which r 0 is not a root. Could r 0 ∗^ be a root of pm? If so, then the argument given at the start of this proof would show that (r 0 ∗)∗^ is also a root of pm. But (r 0 ∗)∗^ = r 0 and we know r 0 is not a root of pm. So it is not possible for r 0 ∗^ to be a root of pm. All this shows that
r 0 is a root of multiplicity m for p(r ) H⇒ r 0 ∗^ is a root of multiplicity m for p(r ).
Replacing r 0 with r 0 ∗^ then gives us
r 0 ∗^ is a root of multiplicity m for p(r ) H⇒
r 0 ∗
is a root of multiplicity m for p(r ).
Together with the fact that (r 0 ∗)∗^ = r 0 , these two implications give us
r 0 is a root of multiplicity m for p(r ) ⇐⇒ r 0 ∗^ is a root of multiplicity m for p(r ) ,
completing our proof of theorem 18.3.
Additional Exercises
18.1. Using clever factoring of the characteristic polynomials (such as done in example 18. on page 382), find the general solution to each of the following:
394 Arbitrary Homogeneous Linear Equations with Constant Coefficients
18.2. For each of the following differential equations, one or more roots to the corresponding characteristic polynomial can be found by “testing candidates” (as illustrated in example
- a. y(^4 ) − 4 y(^3 ) =
- b. y(^4 ) + 4 y′′ =
- c. y(^4 ) − 34 y′′ + 225 y =
- d. y(^4 ) − 81 y =
- e. y(^4 ) − 18 y′′ + 81 y =
- f. y(^5 ) + 18 y(^3 ) + 81 y′ =
- a. y′′′ − y′′ + y′ − y = 18.7 on page 383). Using this fact, find the general solution to each.
- b. y′′′ − 6 y′′ + 11 y′ − 6 y =
- c. y′′′ − 8 y′′ + 37 y′ − 50 y =
- d. y(^4 ) + 2 y(^3 ) + 10 y′′ + 18 y′ + 9 y =
- a. y′′′ + 4 y′ = 0 with y( 0 ) = 4 , y′( 0 ) = 6 and y′′( 0 ) = 18.3. Find the solution to each of the following initial-value problems:
- b. y′′′ − 6 y′′ + 12 y′ − 8 y = - with y( 0 ) = 5 , y′( 0 ) = 13 and y′′( 0 ) =
- c. y(^4 ) + 26 y′′ + 25 y = - with y( 0 ) = 6 , y′( 0 ) = − 28 , y′′( 0 ) = − 102 and y(^3 )( 0 ) =
- a. y′′′ − 8 y = 18.4. Find the general solution to each of the following:
- b. y(^4 ) + 13 y′′ + 36 y =
- c. y(^6 ) − 3 y(^4 ) + 3 y′′ − y =
- d. y(^6 ) − 2 y(^3 ) + y =
