Constant-
Coefficient
Linear
Differential
Equations
Math 240
Homogeneous
equations
Nonhomog.
equations
Constant-Coefficient Linear Differential
Equations
Math 240 — Calculus III
Summer 2013, Session II
Monday, August 5, 2013
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Constant-Coefficient Linear Differential Equations, Lecture notes of Differential Equations
Constant-coefficient linear differential equations, including homogeneous and nonhomogeneous equations. It covers topics such as the auxiliary polynomial, polynomial differential operators, linear polynomial differential operators, distinct linear factors, multiple roots, and their solutions. The document also provides examples and proofs to support the concepts discussed.
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CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Constant-Coefficient Linear Differential
Equations
Math 240 — Calculus III
Summer 2013, Session II
Monday, August 5, 2013
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Agenda
- Homogeneous constant-coefficient linear differential equations
- Nonhomogeneous constant-coefficient linear differential equations
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
The auxiliary polynomial
Consider the homogeneous linear differential equation y(n)^ + a 1 y(n−1)^ + · · · + an− 1 y′^ + any = 0 with constant coefficients ai. Expressed as a linear differential operator, the equation is P (D)y = 0, where P (D) = Dn^ + a 1 Dn−^1 + · · · + an− 1 D + an.
Definition
A linear differential operator with constant coefficients, such as P (D), is called a polynomial differential operator. The polynomial P (r) = rn^ + a 1 rn−^1 + · · · + an− 1 r + an is called the auxiliary polynomial, and the equation P (r) = 0 the auxiliary equation.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
The auxiliary polynomial
Example
The equation y′′^ + y′^ − 6 y = 0 has auxiliary polynomial P (r) = r^2 + r − 6.
Examples
Give the auxiliary polynomials for the following equations.
- y′′^ + 2y′^ − 3 y = 0
- (D^2 − 7 D + 24)y = 0
- y′′′^ − 2 y′′^ − 4 y′^ + 8y = 0
r^2 + 2r − 3 r^2 − 7 r + 24 r^3 − 2 r^2 − 4 r + 8
The roots of the auxiliary polynomial will determine the solutions to the differential equation.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Linear polynomial differential operators
In our example, y′′^ + y′^ − 6 y = 0, with auxiliary polynomial P (r) = r^2 + r − 6 , the roots of P (r) are r = 2 and r = − 3. An equivalent statement is that r − 2 and r + 3 are linear factors of P (r). The functions y 1 (x) = e^2 x^ and y 2 (x) = e−^3 x^ are solutions to y′ 1 − 2 y 1 = 0 and y′ 2 + 3y 2 = 0, respectively.
Theorem
The general solution to the linear differential equation y′^ − ay = 0 is y(x) = ceax.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Theorem
Suppose P (D) and Q(D) are polynomial differential operators P (D)y 1 = 0 = Q(D)y 2. If L = P (D)Q(D), then Ly 1 = 0 = Ly 2.
Proof.
P (D)Q(D)y 2 = P (D)
Q(D)y 2
= P (D)0 = 0
P (D)Q(D)y 1 = Q(D)P (D)y 1 = Q(D)
P (D)y 1
= Q(D)0 = 0 Q.E.D.
Example
The theorem implies that, since (D − 2)y 1 = 0 and (D + 3)y 2 = 0, the functions y 1 (x) = e^2 x^ and y 2 (x) = e−^3 x^ are solutions to y′′^ + y′^ − 6 y = (D^2 + D − 6)y = (D − 2)(D + 3)y = 0.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Distinct linear factors
If we can factor the auxiliary polynomial into distinct linear factors, then the solutions from each linear factor will combine to form a fundamental set of solutions.
Example
Determine the general solution to y′′^ − y′^ − 2 y = 0. The auxiliary polynomial is P (r) = r^2 − r − 2 = (r − 2)(r + 1). Its roots are r 1 = 2 and r 2 = − 1. The functions y 1 (x) = e^2 x and y 2 (x) = e−x^ satisfy (D − 2)y 1 = 0 = (D + 1)y 2. Therefore, y 1 and y 2 are solutions to the original equation. Since we have 2 solutions to a 2 nd^ degree equation, they constitute a fundamental set of solutions; the general solution is y(x) = c 1 e^2 x^ + c 2 e−x.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Multiple roots
What can go wrong with this process? The auxiliary polynomial could have a multiple root. In this case, we would get one solution from that root, but not enough to form the general solution. Fortunately, there are more.
Theorem
The differential equation (D − r)my = 0 has the following m linearly independent solutions: erx, xerx, x^2 erx,... , xm−^1 erx.
Proof.
Check it. Q.E.D.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Complex roots
What happens if the auxiliary polynomial has complex roots? Can we recover real solutions? Yes!
Theorem
If P (D)y = 0 is a linear differential equation with real constant coefficients and (D − r)m^ is a factor of P (D) with r = a + bi and b 6 = 0, then
- P (D) must also have the factor (D − r)m,
- this factor contributes the complex solutions e(a±bi)x, xe(a±bi)x,... , xm−^1 e(a±bi)x,
- the real and imaginary parts of the complex solutions are linearly independent real solutions xkeax^ cos bx and xkeax^ sin bx for k = 0, 1 ,... , m − 1.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Complex roots
Example
Determine the general solution to y′′^ + 6y′^ + 25y = 0.
- The auxiliary polynomial is r^2 + 6r + 25.
- Its has roots r = − 3 ± 4 i.
- Two independent real-valued solutions are y 1 (x) = e−^3 x^ cos 4x and y 2 (x) = e−^3 x^ sin 4x.
- The general solution is y(x) = e−^3 x(c 1 cos 4x + c 2 sin 4x).
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Overview
The technique proceeds from the observation that, if we know a polynomial differential operator A(D) so that A(D)F = 0, then applying A(D) to the nonhomogeneous equation P (D)y = F (1) yields the homogeneous equation A(D)P (D)y = 0. (2) A particular solution to (1) will be a solution to (2) that is not a solution to the associated homogeneous equation P (D)y = 0.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Example
Determine the general solution to (D + 1)(D − 1)y = 16e^3 x.
- The associated homogeneous equation is (D + 1)(D − 1)y = 0. It has the general solution yc(x) = c 1 ex^ + c 2 e−x.
- Recognize the nonhomogeneous term F (x) = 16e^3 x^ as a solution to the equation (D − 3)y = 0.
- The differential equation (D − 3)(D + 1)(D − 1)y = 0 has the general solution y(x) = c 1 ex^ + c 2 e−x^ + c 3 e^3 x.
- Pick the trial solution yp(x) = c 3 e^3 x. Substituting it into the original equation forces us to choose c 3 = 2.
- Thus, the general solution is y(x) = yc(x) + yp(x) = c 1 ex^ + c 2 e−x^ + 2e^3 x.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Finding annihilators
Functions that can be annihilated by polynomial differential operators are exactly those that can arise as solutions to constant-coefficient homogeneous linear differential equations. We have seen that these functions are
- F (x) = cxkeax,
- F (x) = cxkeax^ sin bx,
- F (x) = cxkeax^ cos bx,
- linear combinations of 1–3. If the nonhomogeneous term is one of 1–3, then it can be annihilated by something of the form A(D) = (D − r)k+1, with r = a in 1 and r = a + bi in 2 and 3. Otherwise, annihilators can be found by taking successive derivatives of F and looking for linear dependencies.
CoefficientLinear Differential Equations Math 240
Homogeneous equations Nonhomog. equations
Example
Determine the general solution to (D − 4)(D + 1)y = 16xe^3 x.
- The general solution to the associated homogeneous equation (D − 4)(D + 1)y = 0 is yc(x) = c 1 e^4 x^ + c 2 e−x.
- An annihilator for 16 xe^3 x^ is A(D) = (D − 3)^2.
- The general solution to (D − 3)^2 (D − 4)(D + 1)y = 0 includes yc and the terms c 3 e^3 x^ and c 4 xe^3 x.
- Using the trial solution yp(x) = c 3 e^3 x^ + c 4 xe^3 x, we find the values c 3 = − 3 and c 4 = − 4.
- The general solution is y(x) = yc(x) + yp(x) = c 1 e^4 x^ + c 2 e−x^ − 3 e^3 x^ − 4 xe^3 x.