Example. Mathematics 255: Lecture 17. Example. Example (cont d) Consider the equation. d 2 y dt 2 + dy

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1 Mathematics 255: Lecture 17 Undetermined Coefficients Dan Sloughter Furman University October 10, y = 5e 4t. so the general solution of 0 = r 2 + r 6 = (r + 3)(r 2), 6y = 0 y(t) = c 1 e 3t + c 2 e 2t. Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 For a particular solution of the nonhomogeneous equation, we will guess ψ(t) = Ae 4t for some constant A. ψ (t) = 4Ae 4t ψ (t) = 16e 4t, so we want to find A such that 5e 4t = 16Ae 4t 4Ae 4t 6Ae 4t = 6Ae 4t. Hence A = 5 6, so ψ(t) = 5 6 e 4t. So the general solution of the nonhomogeneous equation y(t) = c 1 e 3t + c 2 e 2t e 4t. 6y = e 3t. In th case, ψ(t) = Ae 3t a solution of the homogeneous equation 6y = 0, so would not work as a guess for a particular solution to the nonhomogeneous equation. Instead, we will guess ψ(t) = Ate 3t for some constant A. Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20

2 Theorem ψ (t) = 3Ate 3t + Ae 3t = (1 3t)Ae 3t ψ (t) = 3(1 3t)Ae 3t 3Ae 3t = (9t 6)Ae 3t. Suppose ψ 1 (t) a solution of the second-order linear equation 2 + p(t)dy + q(t)y = g(t) ψ 2 (t) a solution of the second-order linear equation So we want e 3t = (9t 6)Ae 3t + (1 3t)Ae 3t 6Ate 3t = 5Ae 3t. Hence A = 1 5, so the general solution of the nonhomogeneous equation y(t) = c 1 e 3t + c 2 e 2t 1 5 te 3t 2 + p(t)dy + q(t)y = f (t). ψ(t) = ψ 1 (t) + ψ 2 (t) a solution of 2 + p(t)dy Th the principle of superposition + q(t)y = g(t) + f (t). Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Proof Let L[y](t) = 2 + p(t)dy + q(t)y. L[ψ 1 + ψ 2 ](t) = L[ψ 1 ](t) + L[ψ 2 ](t) = g(t) + f (t). The general solution of 6y = 5e 4t + e 3t y(t) = c 1 e 3t + c 2 e 2t e 4t 1 5 te 3t. Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20

3 Undetermined coefficients Undetermined coefficients (cont d) Consider the second-order, linear, constant-coefficient, nonhomogeneous equation a 2 + b dy + cy = g(t), with corresponding homogeneous equation a 2 + b dy Judicious guesses - g(t) = e αt : + cy = 0. If α not a root of the charactertic equation, let ψ(t) = Ae αt. If α a root of the charactertic equation, but not a double root, let ψ(t) = Ate αt. If α a double root of the charactertic equation, let ψ(t) = At 2 e αt. Judicious guesses - g(t) = cos(ωt) or g(t) = sin(ωt): If cos(ωt) sin(ωt) are not solutions of the homogeneous equation, let ψ(t) = A cos(ωt) + B sin(ωt). If cos(ωt) sin(ωt) are solutions of the homogeneous equation, let ψ(t) = At cos(ωt) + Bt sin(ωt). Judicious guesses - g(t) an nth degree polynomial: If c 0, let If c = 0, let ψ(t) = A 0 + A 1 t + + A n t n. ψ(t) = A 1 t + A 2 t A n+1 t n+1. Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 2 dy 2y = 4t2. 2 dy 2y = 0 0 = r 2 r 2 = (r 2)(r + 1). y(t) = c 1 e 2t + c 2 e t. For a particular solution to the nonhomogeneous equation, we guess ψ(t) = A + Bt + Ct 2. ψ (t) = B + 2Ct ψ (t) = 2C, so we want 4t 2 = 2C (B + 2Ct) 2(A + Bt + Ct 2 ) = (2C B 2A) (2B + 2C)t 2Ct 2. Thus, equating coefficients, we need 4 = 2C 0 = 2B + 2C 0 = 2A B + 2C. It follows that A = 3, B = 2, C = 2. Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20

4 Hence ψ(t) = 3 + 2t 2t 2. And so the general solution of the nonhomogeneous equation y(t) = c 1 e 2t + c 2 e t 3 + 2t 2t y = sin(t). 0 = r y = 0 y(t) = c 1 cos(t) + c 2 sin(t). Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 For a particular solution of the nonhomogeneous equation, we guess We need to find A B such that ψ(t) = At cos(t) + Bt sin(t). sin(t) = (2A + Bt) sin(t) + (2B At) cos(t) + At cos(t) + Bt sin(t) = 2A sin(t) + 2B cos(t). ψ (t) = At sin(t) + A cos(t) + Bt cos(t) + B sin(t) = ( At + B) sin(t) + (Bt + A) cos(t). ψ (t) = ( At + B) cos(t) A sin(t) (Bt + A) sin(t) + B cos(t) = (2A + Bt) sin(t) + (2B At) cos(t). Equating coefficients, we want 2A = 1 2B = 0; that, A = 1 2 B = 0. Thus ψ(t) = 1 2 t cos(t), the general solution of the nonhomogeneous equation y(t) = c 1 cos(t) + c 2 sin(t) 1 2 t cos(t). Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20

5 2 2dy + y = tet. 0 = r 2 2r + r = (r 1) 2 2 2dy + y = 0 y(t) = c 1 e t + c 2 te t. If r = 1 were not a root of the charactertic equation, we would guess ψ(t) = (A + Bt)e t. If r = 1 were a root, but not a double root, of the charactertic equation, we would guess ψ(t) = t(a + Bt)e t. Since r = 1 a double root of the charactertic equation, we guess ψ(t) = t 2 (A + Bt)e t = (At 2 + Bt 3 )e t. Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 ψ (t) = (At 2 +Bt 3 )e t +(2At +3Bt 2 )e t = (2At +(A+3B)t 2 +Bt 3 )e t ψ (t) = (2At + (A + 3B)t 2 + Bt 3 )e t + (2A + 2(A + 3B)t + 3Bt 2 )e t = (2A + (4A + 6B)t + (A + 6B)t 2 + Bt 3 )e t. Hence we want te t = (2A + (4A + 6B)t + (A + 6B)t 2 + Bt 3 )e t 2(2At + (A + 3B)t 2 + Bt 3 )e t + (At 2 + Bt 3 )e t = (2A + 6Bt)e t Equating coefficients, we need 0 = 2A 1 = 6B. Thus B = 1 6, so ψ(t) = 1 6 t3 e t. Thus the general solution of the nonhomogeneous equation y(t) = c 1 e t + c 2 te t t3 e t. Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20 Dan Sloughter (Furman University) Mathematics 255: Lecture 17 October 10, / 20