Section 3.4. Second Order Nonhomogeneous Equations y + p(x)y + q(x)y = f(x) (N) The corresponding homogeneous equation y + p(x)y + q(x)y = 0 (H) is called the reduced equation of (N). 1
General Results THEOREM 1: If z = z 1 (x) and z = z 2 (x) are solutions of equation (N), then y(x) = z 1 (x) z 2 (x) is a solution of equation (H). Proof: 2
THEOREM 2: Let y = y 1 (x) and y = y 2 (x) be linearly independent solutions of the reduced equation (H) and let z = z(x) be a particular solution of (N). Then y(x) = C 1 y 1 (x) + C 2 y 2 (x) + z(x) is the general solution of (N). 3
The general solution of (N) consists of the general solution of the reduced equation (H) plus a particular solution of (N): y = C 1 y 1 (x) + C 2 y 2 (x) }{{} + z(x). }{{} gen soln (H) + part soln (N) 4
To find the general solution of (N) you need to find: (i) a fundamental set of solutions y 1, y 2 of the reduced equation (H), and (ii) a particular solution z of (N). 5
THEOREM 3: (Superposition Principle) Given the nonhomogeneous equation y + p(x)y + q(x)y = f(x) + g(x). If z = z f (x) is a particular solution of y + p(x)y + q(x)y = f(x) and z = z g (x) is a particular solution of y + p(x)y + q(x)y = g(x), 6
then z(x) = z f (x) + z g (x) is a particular solution of y + p(x)y + q(x)y = f(x) + g(x). Proof:
Variation of Parameters Let y = y 1 (x) and y = y 2 (x) be independent solutions of the reduced equation (H) and let W(x) = y 1 y 2 y 1 y 2 = y 1y 2 y 2y 1 0 be their Wronskian. Then y = C 1 y 1 (x) + C 2 y 2 (x) is the general solution of (H). 7
Set z(x) = u(x)y 1 (x) + v(x)y 2 (x) where u and v are to be determined so that z is a solution of (N). z = uy 1 + vy 2 z = uy 1 + y 1u + vy 2 + y 2v Set y 1 u + y 2 v = 0 so we have z = uy 1 + vy 2 z = uy 1 + vy 2 and z = uy 1 + y 1 u + vy 2 + y 2 v
Substitute z, z, z into the differential equation ( uy 1 + y 1 u + vy 2 + y 2 v ) +p ( uy 1 + ) vy 2 + q (uy 1 + vy 2 ) = f Rearrange the terms: u ( y 1 + py 1 + qy ) ( 1 + v y 2 + py 2 + qy ) 2 + y 1 u + y 2 v = f 8
We now have two equations in the two unknowns u and v : y 1 u + y 2 v = 0 y 1u + y 2v = f Solve for u : ( y1 y 2 y 2y 1 ) u = y 2 f or Wu = y 2 f so u = y 2f W and u = y 2 f W dx; 9
y 1 u + y 2 v = 0 y 1u + y 2v = f Solve for v : ( y1 y 2 y 2y 1 ) v = y 1 f or Wv = y 1 f so v = y 1f W and v = y 1 f W dx; 10
Therefore, z(x) = y 1 (x) y 2 (x)f(x) W(x) dx+y 2 (x) y 1 (x)f(x) W(x) dx is a particular solution of the nonhomogeneous equation (N). 11
Examples: 1. {y 1 (x) = x 2, y 2 (x) = x 4 } is a fundamental set of solutions of y 5 x y + 8 x 2 y = 0. Find a particular solution z of the equation. y 5 x y + 8 x 2 y = 4x3 12
y 1 = x 2, y 2 = x 4 W[y 1, y 2 ] = 13
2. Find the general solution of y 4y + 4y = e2x x 14
3. Find the general solution of y + y 6y = 3e 2x 15
4. Find the general solution of y + 4y = 2tan2x 16
Section 3.5. Undetermined Coefficients aka Guessing NOTE: This method can be used only when: 1. the de has constant coefficients 2. f is an exponential function y + ay + by = f(x) a, b constants, f an exponential function. 17
Case 1: If y + ay + by = ce rx Set z(x) = Ae rx Example 1: Find a particular solution of y 5y + 6y = 7e 4x. Set z = Ae 4x where A is to be determined. 18
Answer: z = 1 6 e 4x. The general solution of the differential equation is: y = C 1 e 2x + C 2 e 3x + 1 6 e 4x. Note: If L[y] = y + ay + by, then L[Ae rx ] = A ( r 2 + ar + b ) e rx = Ke rx That is, L[Ae rx ] is a constant multiple of e rx 19
Example 2: Find a particular solution of y + 2y 3y = 9e 2x. 20
Case 2: y + ay + by = c cos βx, or y + ay + by = c sinβx, or y +ay +by = c cos βx+d sinβx, Example: y 2y + y = 5cos 2x 21
Note: If L[y] = y + ay + by, then L[Acos βx] = K cos βx + M sin βx That is, L[Acos βx] involves BOTH cosine and sine. Similarly for L[B sin βx] and L[Acos βx + B sinβx] Therefore, if f(x) = ccos βx or f(x) = dsin βx or f(x) = c cos βx + d sin βx 22
set z(x) = A cos βx + B sin βx where A, B are to be determined 23
Example 3: Find a particular solution of y 2y + y = 5cos 2x. Set z = Acos 2x + B sin 2x 24
Answer: z = 3 5 cos 2x 4 5 sin 2x. The general solution of the differential equation is: y = C 1 e x + C 2 xe x 3 5 cos 2x 4 5 sin 2x. 25
Example 4: Find a particular solution of y 2y + 5y = 2cos 3x 4sin 3x Set z = Acos 3x + B sin 3x + Ce 2x where A, B, C are to be determined. 26
y 2y + 5y = 2cos 3x 4sin 3x 27
Case 3: If f(x) = ce αx cos βx, de αx sin βx or ce αx cos βx + de αx sin βx set z(x) = Ae αx cos βx + Be αx sin βx where A, B are to be determined. 28
Example 5: Find a particular solution of y + 9y = 4e x sin 2x. Set z = Ae x cos 2x + Be x sin 2x Answer: z = 4 13 ex cos 2x+ 6 13 ex sin 2x. 29
Example 6: Find a particular solution z of y 2y + y = 3e 3x 5sin2x. 30
Example 7: Find a particular solution z of y + y 6y = 3e 2x. 31
A BIG Difficulty: The trial solution z is a solution of the reduced equation. In this case, y 1 = e 3x and y 2 = e 2x are solutions of the reduced equation y + y 6y = 0 From Example 3, Section 3.4: z = 3 5 x e2x 32
Example 8: Find a particular solution z of y + y 6y = 3e 2x. Set z = Ae 2x? NO this satisfies the reduced equation, Set z = Axe 2x 33
Example 9: Find a particular solution of y 2y 15y = 6e 3x 34
Example 10: Find a particular solution z of y + 4y = 2cos 2x. 35
Example 11: Find a particular solution z of y + 6y + 9y = 4e 3x. 36
Example 12: Find a particular solution z of y 2y 8y = 3e 2x + 6 37
Example 13: Find a particular solution z of y 3y = 4e 3x + 2 38
Answers: z 9 = 3 4 x e 3x z 10 = 1 2 xsin 2x z 11 = 2x 2 e 3x z 12 = 1 2 xe 2x 3 4 z 13 = 4 3 x e3x 2 3 x 39
The Method of Undetermined Coefficients A. Applies only to equations of the form y + ay + by = f(x) where a, b are constants and f is an exponential function. c.f Variation of Parameters 40
B. Basic Case: If: f(x) = ae rx set z = Ae rx. f(x) = ccos βx, dsin βx, or ccos βx + dsin βx, set z = Acos βx + B sin βx. f(x) = ce αx cos βx, de αx sin βx or ce αx cos βx + de αx sin βx, set z = Ae αx cos βx + Be αx sin βx. 41
BUT: If z satisfies the reduced equation, use xz; if xz also satisfies the reduced equation, then x 2 z will give a particular solution. 42
C. General Case: If f(x) = p(x)e rx where p is a polynomial of degree n, then set z = P(x)e rx where P is a polynomial of degree n with undetermined coefficients. 43
Example 14: Find a particular solution of y 2y 8y = (4x + 5)e 2x. Set z = (Ax + B)e 2x. 44
Example 15: Find a particular solution of y 3y + 2y = (2x 2 1)e x. Set z = (Ax 2 + Bx + C)e x. Answer: z = ( 1 3 x 2 + 9 5 x + 27 5 ) e x. 45
If f(x) = p(x)cos βx + q(x)sin βx where p, q are polynomials, then set z = P(x)cos βx + Q(x)sin βx where P, Q are polynomials of degree n with undetermined coefficients, n = max degree of p and q. 46
Example 16: y 2y 3y = 3 cos x + (x 2)sin x. Set z = (Ax+B)cos x+(cx+d)sin x. 47
Answer: z = ( 1 10 x 47 ) ( 1 cos x 50 5 x 2 ) sin x. 25 48
If f(x) = p(x)e αx cos βx + q(x)e αx sin βx where p, q are polynomials of degree n, then set z = P(x)e αx cos βx + Q(x)e αx sin βx where P, Q are polynomials of degree n with undetermined coefficients. 49
Example 17: y +4y = 2x e x cos x. Set z = (Ax+B)e x cos x+(cx+d)e x sin x Answer: z = 1 25 (10x 7)ex cos x+ 1 25 (5x 1)ex sin x. 50
Example 18: Find a particular solution of y 2y 8y = (4x + 5)e 2x. Set z = (Ax + B)e 2x???? 51
BUT: Warning!!! If any part of z satisfies the reduced equation, try xz; if any part of xz also satisfies the reduced equation, then x 2 z will give a particular solution. 52
Examples: 1. Give the form of a particular solution of y 4y 5y = 2cos 3x 5e 5x + 4. Answer: z = Acos 3x + B sin 3x + Cxe 5x + D 53
Answer: z = Ax + B + Cx 2 e 4x 54 2. Give the form of a particular solution of y + 8y + 16y = 2x 1 + 7e 4x.
C cos 2x + Dsin 2x + Ee 2x 55 3. Give the form of a particular solution of y + y = 4sin x cos 2x + 2e 2x. Answer: z = Axcos x + Bxsin x +
4. Give the form of the general solution of y + 9y = 4cos 2x + 3sin 2x Answer: y = C 1 cos 3x + C 2 sin 3x + Acos 2x + B sin 2x 56
5. Give the form of a particular solution of y + 9y = 4cos 3x + 3sin 2x Answer: z = Axcos 3x + Bxsin 3x + C cos 2x + Dsin 2x 57
6. Give the form of the general solution of y + 4y + 4y = 4xe 2x + 3 Answer: y = C 1 e 2x + C 2 xe 2x + (Ax 3 + Bx 2 )e 2x + C 58
7. Give the form of the general solution of y + 4y + 4y = 4e 2x sin 2x + 3x Answer: y = C 1 e 2x + C 2 xe 2x + Ae 2x cos 2x + Be 2x sin 2x + Cx + D 59
8. Give the form of the general solution of y + 4y = 4sin 2x + 3 Answer: y = C 1 e 4x +C 2 +Acos 2x+B sin 2x+Cx 60
9. Give the form of a particular solution of y + 2y + 10y = 2e 3x sin x + 4e 3x Answer: z = Ae 3x cos x + Be 3x sin x + Ce 3x 61
10. Give the form of the general solution of y + 2y + 10y = 2e x sin 3x + 2e x Answer: y = C 1 e x cos 3x+C 2 e x sin 3x+ Axe x cos 3x + Bxe x sin 3x + Ce x 62
11. Give the form of a particular solution of y 2y 8y = 2cos 3x (3x+1)e 2x 4 Answer: z = Acos 3x+B sin 3x+(Cx 2 +Dx)e 2x +E 63
12. Give the form of a particular solution of y 2y 8y = 2cos 3x 3xe 2x 3x Answer: z = Acos 3x+B sin 3x+(Cx 2 +Dx)e 2x +Ex + F 64
13. Find the general solution of y 4y + 4y = 4e 2x + e2x x Answer: y = C 1 e 2x + C 2 xe 2x 2x 2 e 2x + xe 2x ln x 65
Summary: 1. Variation of parameters: Can be applied to any linear nonhomogeneous equations, but requires a fundamental set of solutions of the reduced equation. 66
2. Undetermined coefficients: Is limited to linear nonhomogeneous equations with constant coefficients, and f must be an exponential function, f(x) = ae rx, f(x) = ccos βx + dsin βx, f(x) = ce αx cos βx + de αx sin βx, or p(x)f(x) p a polynomial. 67
In cases where both methods are applicable, the method of undetermined coefficients is usually more efficient and, hence, the preferable method. 68
Section 3.6. Vibrating Mechanical Systems 69
I. Free Vibrations Hooke s Law: The restoring force of a spring is proportional to the displacement: F = ky, k > 0. Newton s Second Law: Force equals mass times acceleration: F = ma = m d2 y dt 2. Mathematical model: m d2 y dt 2 = ky 70
which can be written where ω = k/m. d 2 y dt 2 + ω2 y = 0 ω is called the natural frequency of the system.
The general solution of this equation is: y = C 1 sin ωt + C 2 cos ωt which can be written y = Asin(ωt + φ). A is called the amplitude, φ is called the phase shift. 71
y = C 1 sin ωt + C 2 cos ωt 72
y = Asin(ωt + φ) A y t A 73
II. Forced Free Vibrations Apply an external force G to the freely vibrating system Force Equation: F = ky + G. Mathematical Model: my = ky+g or y + k m y = G m, a nonhomogeneous equation. 74
A periodic external force: G = a cos γt, a, γ > 0 const. Force Equation: F = ky + a cos γt Mathematical Model: y + k m y = a m cos γt y + ω 2 y = α cos γt where ω = k/m, α = a m. 75
ω is called the natural frequency of the system, γ is called the applied frequency. 76
Case 1: γ ω. y + ω 2 y = α cos γt General solution, reduced equation: y = C 1 cos ωt+c 2 sin ωt = Asin(ωt+φ 0 ). Form of particular solution (undetermined coefficients): z = Acos γt + B sin γt. A particular solution: z = α ω 2 γ 2 cos γt. 77
General solution: y = Asin(ωt + φ 0 ) + α ω 2 cos γt. γ2
ω/γ rational: periodic motion 1 ω/γ irrational: not periodic y 1 t 78
Case 2: γ = ω. y + ω 2 y = αcos ωt General solution, reduced equation: y = C 1 cos ωt + C 2 sin ωt = Asin(ωt + φ 0 ). Form of particular solution (undetermined coefficients): z = A tcos γt + B tsin γt. 79
A particular solution: α 2ω t sin ωt. General solution: y = Asin(ωt + φ 0 ) + α 2ω t sin ωt. Unbounded oscillation This is known as resonance 80
y = Asin(ωt + φ 0 ) + α 2ω t sin ωt. y t 81
III. Damped Free Vibrations: A resistance force R (e.g., friction) proportional to the velocity v = y and acting in a direction opposite to the motion: R = cy with c > 0. Force Equation: F = ky(t) cy (t). Newton s Second Law: F = ma = my 82
Mathematical Model: my (t) = ky(t) cy (t) or y + c m y + k m y = 0 (c, k, m constant) or y + αy + βy = 0 α = c m, β = k m α, β positive constants.
Characteristic equation: r 2 + α r + β = 0. Roots r = α ± α 2 4β. 2 There are three cases to consider: α 2 4β < 0, α 2 4β > 0, α 2 4β = 0. 83
Case 1: α 2 4β < 0. Complex roots: (Underdamped) r 1 = α 2 + iω, r 2 = α 2 iω where ω = 4β α 2. 2 General solution: y = e ( α/2)t (C 1 cos ωt + C 2 sin ωt) 84
or y(t) = A e ( α/2)t sin(ωt + φ 0 ) where A and φ 0 are constants, NOTE: The motion is oscillatory AND y(t) 0 as t. 85
Underdamped Case: y t 86
Case 2: α 2 4β > 0. Two distinct real roots: (Overdamped) r 1 = α + α 2 4β, r 2 = α α 2 4β. 2 2 General solution: y(t) = y = C 1 e r 1t + C 2 e r 2t. The motion is nonoscillatory. 87
NOTE: Since α 2 4β < α 2 = α, r 1 and r 2 are both negative and y(t) 0 as t. 88
Case 3: α 2 4β = 0. One real root: (Critically Damped) r 1 = r 2 = α 2, General solution: y(t) = y = C 1 e (α/2) t + C 2 t e (α/2) t. The motion is nonoscillatory and y(t) 0 as t. 89
Overdamped and Critically Damped Cases: y y t t y t 90
Summary of Case III: All solutions of y + ay + by = 0 have limit 0 as t. That is, in the presence of a resistant force (friction), all solutions ultimately return to the equilibrium position. 91
IV. Forced Damped Vibrations Apply an external force G to a damped, freely vibrating system Force Equation: F = ky cy + G. Mathematical Model: my = ky cy + G or y + c m y + k m y = G m, 92
which we write as y + αy + βy = g, where α = c/m, β = k/m, g = G/m A periodic external force: g = a cos γt, a, γ > 0 const. Mathematical Model: y + αy + βy = acos γt 93
General solution: y(t) = C 1 y 1 (t) + C 2 y 2 (t) + Z(t) = Y c (t) + Z(t), Note: From Case III lim t Y c(t) = 0. as t so lim t y(t) = Z(t). 94
Particular solution of (N) y + α y + β y = a cos γt will have the form: Z(t) = Acos γt + B sin γt. General solution of (N): y(t) = Y c (t) + Z(t) Note: lim y(t) = Z(t) t 95
Y c (t), the general solution of the reduced equation, is called the transient solution. Z(t) the particular solution of (N), is called the steady state solution. 96
Example 1: y + 3 4 y + 1 8 y = cos t General solution y = C 1 e t/4 +C 2 e t/2 + 56 85 cos t 48 85 sin t Transient solution: y(t) = 2e t/4 +e t/2 (C 1 = 2, C 2 = 1) Steady-state solution: Z(t) = 56 85 cos t 48 85 sin t 97
Transient solution: Steady-state solution: - 98
y = 2e t/4 + e t/2 + 56 85 cos t 48 85 sin t - 99
Example 2: y + 2y + 5y = cos t General solution y = C 1 e t cos 2t + C 2 e t sin 2t+ Transient solution: 1 5 cos t + 1 10 sin t y(t) = 2e t cos2t (C 1 = 2, C 2 = 0) Steady-State solution: Z(t) = 1 5 cos t + 1 10 sin t 100
Transient solution. Steady-State solution: 101
y = 2e t cos2t + 1 5 cos t + 1 10 sin t 102