Chapter 6: Non-homogeneous Equations: Undetermined Coefficients MATH 374 Lecture 18 Chapter 6: Non-homogeneous Equations: Undetermined Coefficients

6.1: Construction of a Homogeneous Equation from a Specified Solution In Section 4.6 of our class notes, we saw that the general solution to the equation (b0 Dn + b1 Dn-1 + … + bn-1 D + bn)y = R(x) (1) is y = yc + yp where yc (complementary function) is the general solution of the homogeneous equation (b0 Dn + b1 Dn-1 + … + bn-1 D + bn)y = 0 (2) and yp is a particular solution of (1). 2

A way to find yp One method to find yp when b0, b1, … , bn are constants is the Method of Undetermined Coefficients. In preparation for this method, we will first look at how to construct a homogeneous linear equation of which a given function is a solution. 3

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) 4

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax 5

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a 6

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a 7

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax 8

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a 9

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 10

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax 11

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a 12

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 13

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax 14

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) 15

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) (D-a)n 16

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) (D-a)n c5 eax cos bx 17

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) (D-a)n c5 eax cos bx m = a + bi, a - bi 18

Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) (D-a)n c5 eax cos bx m = a + bi, a - bi (D-a)2 + b2 or 19

[D-(a+bi)][D-(a-bi)] Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) (D-a)n c5 eax cos bx m = a + bi, a - bi (D-a)2 + b2 or [D-(a+bi)][D-(a-bi)] 20

[D-(a+bi)][D-(a-bi)] Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) (D-a)n c5 eax cos bx m = a + bi, a - bi (D-a)2 + b2 or c6 eax sin bx [D-(a+bi)][D-(a-bi)] 21

[D-(a+bi)][D-(a-bi)] Types of Solutions Recall that for homogeneous linear equations with constant coefficients, the following types of solutions occur: Solution Root of f(m) = 0 Factor of f(D) c1 eax m = a D-a c2 x eax m = a, a (D-a)2 c3 x2 eax m = a, a, a (D-a)3 c4 xn-1 eax m = a, a, … , a (n a’s) (D-a)n c5 eax cos bx m = a + bi, a - bi (D-a)2 + b2 or c6 eax sin bx [D-(a+bi)][D-(a-bi)] 22

Example 1 Find a homogeneous linear equation with real constant coefficients that is satisfied by y = 7 + 2x e4x – 5 e6x sin x. Solution: Find the roots of f(m) = 0. 7  m = 0 2x e4x  m = 4, 4 -5 e6x sin x  m = 6 § i Hence an auxiliary equation that has these roots is m(m-4)2((m-6)2 + 12) = 0. Therefore, a corresponding differential equation is D(D-4)2((D-6)2 + 12)y = 0. Multiplying out the operator, we find: (D5 – 20 D4 + 149 D3 -488 D2 + 592 D)y = 0. 23

Example 2 24 Repeat Example 1 with y = 3 sin 3x + x sin 2x. Solution: 3 sin 3x  m = § 3i x sin 2x  m = § 2i, § 2i. ) an auxiliary equation is (m2 + 9) (m2 + 4)2 = 0. ) a desired differential equation is (D2 + 9)(D2 + 4)2 y = 0 or (D6 + 17 D4 + 88 D2 + 144) y = 0. 24

6.2: The Method of Undetermined Coefficients We now use the ideas from Section 6.1 to help us solve the non-homogeneous problem f(D)y = R(x) (1) where f(D) is a differential operator with constant coefficients. The Method of Undetermined Coefficients is outlined on the following slides! 25

Method of Undetermined Coefficients f(D)y = R(x) (1) Method of Undetermined Coefficients Find the roots of the auxiliary equation f(m) = 0: m = m1, m2, … , mn. (2) The general solution to (1) will be y = yc + yp (3) where yc is the complementary function and yp is a particular solution to be determined. 26

Method of Undetermined Coefficients f(D)y = R(x) (1) Method of Undetermined Coefficients If possible, find a homogeneous linear differential equation with constant coefficients that R(x) solves, namely g(D)R = 0, (4) whose auxiliary equation g(m) = 0 has roots m’ = m’1, m’2, … , m’k, (5) found by inspection of R(x) (see Section 6.1). 27

Method of Undetermined Coefficients f(D)y = R(x) (1) Method of Undetermined Coefficients Form the differential equation g(D)f(D)y = 0. (6) From (2) and (5), this differential equation has as roots of its auxiliary equation, g(m)f(m) = 0, m = m1, m2, … , mn m’ = m’1, m’2, … , m’k. The general solution of (6) is of the form y = yc + yq. (7) 28

Method of Undetermined Coefficients f(D)y = R(x) (1) g(D)f(D)y = 0. (6) Method of Undetermined Coefficients Note: Any particular solution yp of (1) must be a solution of (6). This follows since f(D)yp = R(x) ) g(D)f(D)yp = g(D)R = 0. (by (4), due to our choice of g(D)) ) we can write yp = Yc + Yq by (7), for some choice of coefficients in yc and yq. Now, f(D)yp = R(x) ) f(D)(Yc + Yq) = R(x) ) f(D)Yc + f(D)Yq = R(x) ) f(D)Yq = R(x) (since f(D)Yc = 0) (8) ), some particular solution of (1) comes from an appropriate choice of coefficients of yq!!. The correct choice of coefficients is determined by equating coefficients in (8). The general solution of form (3) follows with the yp found. 29

Remarks Not every yq will solve (1). (HW – show this!) f(D)y = R(x) (1) Remarks Not every yq will solve (1). (HW – show this!) The Method of Undetermined Coefficients is applicable if and only if the RHS of (1) is itself a particular solution of some homogeneous linear differential equation with constant coefficients. (See B&D pp. 182 – 184 for a proof that this method works in the second order case.) 30

Example 1: Solve: (D2 – 4) y = e2x + 2. Solution: Use the Method of Undetermined Coefficients! (D2 – 4) y = 0 ) f(D) = (D2 – 4), so the roots of auxiliary equation f(m) = 0, i.e. m2 – 4 = 0, are m = § 2. Hence yc = c1 e-2x + c2 e2x. Find g(D) so that g(D)R = 0, i.e. g(D)(e2x + 2) = 0. Roots of the auxiliary equation g(m) = 0 are m’ = 2, 0. ) g(D) = D(D-2). g(D)f(D)y = 0 ) D(D-2)(D2 – 4)y = 0 ) D(D-2)2(D+2)y = 0 ) y = c1 e-2x + c2 e2x + c3 x e2x + c4 yc yq 31

Example 1: Solve: (D2 – 4) y = e2x + 2. Let yp = Ax e2x + B. Substitute yp into our original differential equation to find A and B. (D2 – 4)(Axe2x + B) = e2x + 2 ) D[D(Axe2x + B)] – 4(Axe2x + B) = e2x + 2 ) D[Ae2x + 2Axe2x] – 4Axe2x – 4B = e2x + 2 ) 2Ae2x + 2Ae2x + 4Axe2x – 4Axe2x – 4B = e2x + 2 ) 4Ae2x – 4B = e2x + 2 ) 4A = 1 and -4B = 2 must hold, since e2x and 1 are linearly independent. ) A = 1/4 and B = -1/2. It follows that the general solution is y = yc + yp = c1 e-2x + c2 e2x + 1/4 x e2x – 1/2. 32

Example 2: Solve y’’’ – y’’ + y’ -y = 4 sin x. Solution: Use the Method of Undetermined Coefficients! (D3 – D2 + D -1)y = 0 ) auxiliary equation f(m) = 0 is m3 – m2 + m – 1 = 0 ) roots are m = 1, § i. (To see this, factor by grouping, to get m2(m-1) + 1(m-1) = 0 ) (m-1)(m2+1) = 0. Or use synthetic division.) Thus yc = c1 ex + c2 cos x + c3 sin x.

Example 2: Solve y’’’ – y’’ + y’ -y = 4 sin x. g(D)(4 sin x) = 0 ) roots of auxiliary equation g(m) = 0 are m’ = § i. Therefore, g(D) = (D2 + 1). g(D)f(D)y = 0 ) (D-1)(D2 + 1)2y = 0 ) y = c1 ex + c2 cos x + c3 sin x + c4 x cos x + c5 x sin x yc yq

Example 2: Solve y’’’ – y’’ + y’ -y = 4 sin x. Let yp = Ax cos x + Bx sin x. Then Dyp = Acos x + Ax(-sin x) + B sin x + Bx cos x D2yp = A(-sin x) + A(-sin x) + Ax(-cos x) + B cos x + B cos x + Bx (-sin x) = -2Asin x + 2Bcos x – Ax cos x – Bx sin x D3yp = -2Acos x + 2B(-sin x) – Acos x – Ax(-sin x) – Bsin x – Bx cos x = -3Acos x – 3Bsin x + Ax sin x – Bx cos x

Example 2: Solve y’’’ – y’’ + y’ -y = 4 sin x. ) yp’’’ – yp’’ + yp’ – yp = - 3Acos x – 3Bsin x + Axsin x – Bxcos x - (-2Asin x + 2Bcos x – Axcos x – Bxsin x) + (Acos x + Ax(-sin x) + Bsin x + Bxcos x) - (Axcos x + Bxsin x)

Example 2: Solve y’’’ – y’’ + y’ -y = 4 sin x. ) yp’’’ – yp’’ + yp’ – yp = - 3Acos x – 3Bsin x + Axsin x – Bxcos x - (-2Asin x + 2Bcos x – Axcos x – Bxsin x) + (Acos x + Ax(-sin x) + Bsin x + Bxcos x) - (Axcos x + Bxsin x) = (-2A-2B)cos x + (2A – 2B)sin x = -2(A+B)cos x + 2(A-B)sin x

Example 2: Solve y’’’ – y’’ + y’ -y = 4 sin x. ) yp’’’ – yp’’ + yp’ – yp = 4 sin x ) -2(A+B)cos x + 2(A – B)sin x = 4 sin x ) Since cos x and sin x are linearly independent, A + B = 0 A – B = 2 ) 2A = 2 ) A = 1 and B = -1 Hence the general solution is y = c1 ex + c2 cos x + c3 sin x + x cos x – x sin x.