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). 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: (Superposition Principle) 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), then z(x) = z f (x) + z g (x) 6
is a particular solution of y + p(x)y + q(x)y = f(x) + g(x).
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 be their Wronskian. Set z(x) = y 1 (x)u(x) + y 2 (x)v(x) where u and v are to be determined so that z is a solution of (N). 7
u (x) = y 2(x)f(x), W (x) u(x) = y 2 (x)f(x) W (x) dx; v (x) = y 1(x)f(x) W (x) v(x) = y 1 (x)f(x) W (x) dx and z(x) = y 1 (x) y 2 (x)f(x) W (x) dx + y 2 (x) y 1 (x)f(x) W (x) dx 8
Examples: 1. {y 1 (x) = x 2, y 2 (x) = x 4 } is a fundamental set of solutions of y 5 x y + 8 x2y = 0. Find a particular solution z of the equation. y 5 x y + 8 x2y = 4x3 9
y 1 = x 2, y 2 = x 4 W [y 1, y 2 ] = 10
2. Find the general solution of y 4y + 4y = e2x x 11
3. Find the general solution of y + y 6y = 3e 2x 12
4. Find the general solution of y + 4y = 2 tan 2x 13
Section 3.5. Undetermined Coefficients aka Guessing NOTE: This method can be used only when: 1. the de has constant coefficients 2. f an exponential function y + ay + by = f(x) a, b constants, f an exponential function. 14
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. 15
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 16
Example 2: Find a particular solution of y + 2y 3y = 9e 2x. 17
Case 2: If y + ay + by = c cos βx, or y + ay + by = c sin βx, or y +ay +by = c cos βx+d sin βx, Example: Find a particular solution of y 2y + y = 5 cos 2x. 18
Note: If L[y] = y + ay + by, then L[A cos βx] = K cos βx + M sin βx That is, L[A cos βx] involves BOTH cosine and sine. Similarly for L[B sin βx] and L[A cos βx + B sin βx] Therefore, if f(x) = c cos βx or f(x) = d sin βx or f(x) = c cos βx + d sin βx 19
set z(x) = A cos βx + B sin βx where A, B are to be determined 20
Example 3: Find a particular solution of y 2y + y = 5 cos 2x. Set z = A cos 2x + B sin 2x 21
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. 22
Example 4: Find a particular solution of y 2y + 5y = 2 cos 3x 4 sin 3x + e 2x Set z = A cos 3x + B sin 3x + Ce 2x where A, B, C are to be determined. 23
y 2y + 5y = 2 cos 3x 4 sin 3x + e 2x 24
Answer: z = 8 13 cos 3x + 1 13 sin 3x + 1 5 e2x. 25
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. 26
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. 27
Example 6: Find a particular solution z of y + y 6y = 3e 2x. 28
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 29
Example 7: Find a particular solution z of y + y 6y = 3e 2x. Since z = Ae 2x? NO this satisfies the reduced equation, Set z = Axe 2x 30
Example 8: Find a particular solution of y 2y 15y = 6e 3x 31
Example 9: Find a particular solution z of y + 4y = 2 cos 2x. 32
Example 10: Find a particular solution z of y + 6y + 9y = 4e 3x. 33
Example 11: Find a particular solution z of y 2y 8y = 3e 2x + 6 34
Example 12: Find a particular solution z of y 3y = 4e 3x + 2 35
Answers: z 8 = 3 4 x e 3x z 9 = 1 2 x sin 2x z 10 = 2x 2 e 3x z 11 = 1 2 xe 2x 3 4 z 12 = 4 3 x e3x 2 3 x 36
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 37
B. Basic Case: If: f(x) = ae rx set z = Ae rx. f(x) = c cos βx, d sin βx, or c cos βx + d sin βx, set z = A cos 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. 38
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. 39
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. 40
Example 13: Find a particular solution of y 2y 8y = (4x + 5)e 2x. Set z = (Ax + B)e 2x. 41
Example 14: 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 + 5 9 x + 5 27) e x. 42
If f(x) = p(x) cos βx + q(x) sin βx where p, q are polynomials of degree n, then set z = P (x) cos βx + Q(x) sin βx where P, Q are polynomials of degree n with undetermined coefficients. 43
Example 15: y 2y 3y = 3 cos x + (x 2) sin x. Set z = (Ax+B) cos x+(cx+d) sin x. 44
Answer: z = ( 1 10 x 47 ) ( 1 cos x 50 5 x 2 ) sin x. 25 45
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. 46
Example 16: 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. 47
Example 17: Find a particular solution of y 2y 8y = (4x + 5)e 2x. Set z = (Ax + B)e 2x. 48
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. 49
Examples: 1. Give the form of a particular solution of y 4y 5y = 2 cos 3x 5e 5x + 4. Answer: z = A cos 3x + B sin 3x + Cxe 5x + D 50
Answer: z = Ax + B + Cx 2 e 4x 51 2. Give the form of a particular solution of y + 8y + 16y = 2x 1 + 7e 4x.
C cos 2x + D sin 2x + Ee 2x 52 3. Give the form of a particular solution of y + y = 4 sin x cos 2x + 2e 2x. Answer: z = Ax cos x + Bx sin x +
4. Give the form of the general solution of y + 9y = 4 cos 2x + 3 sin 2x Answer: y = C 1 cos 3x + C 2 sin 3x + A cos 2x + B sin 2x 53
5. Give the form of a particular solution of y + 9y = 4 cos 3x + 3 sin 2x Answer: z = Ax cos 3x + Bx sin 3x + C cos 2x + D sin 2x 54
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 55
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 56
8. Give the form of the general solution of y + 4y = 4 sin 2x + 3 Answer: y = C 1 e 4x +C 2 +A cos 2x+B sin 2x+Cx 57
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 58
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 59
11. Give the form of a particular solution of y 2y 8y = 2 cos 3x (3x+1)e 2x 4 Answer: z = A cos 3x+B sin 3x+(Cx 2 +Dx)e 2x +E 60
12. Give the form of a particular solution of y 2y 8y = 2 cos 3x 3xe 2x 3x Answer: z = A cos 3x+B sin 3x+(Cx 2 +Dx)e 2x +Ex + F 61
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 62
Summary: 1. Variation of parameters: Can be applied to any linear nonhomogeneous equations, but requires a fundamental set of solutions of the reduced equation. 63
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) = c cos βx + d sin βx, f(x) = ce αx cos βx + de αx sin βx, or p(x)f(x) p a polynomial. 64
In cases where both methods are applicable, the method of undetermined coefficients is usually more efficient and, hence, the preferable method. 65
Section 3.6. Vibrating Mechanical Systems undamped (no friction) damped (with friction) free (e.g., only the spring) forcing (add external force) 66
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 67
which can be written d 2 y dt 2 + ω2 y = 0 where ω = k/m. 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 = A sin (ωt + φ). A is called the amplitude, φ is called the phase shift. 68
y = A sin(ωt + φ) 69
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. 70
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. 71
ω is called the natural frequency of the system, γ is called the applied frequency. if omega = gamma, will have a RESONANCE 72
Case 1: γ ω. y + ω 2 y = α cos γt General solution, reduced equation: y = C 1 cos ωt+c 2 sin ωt = A sin (ωt+φ 0 ). Form of particular solution (undetermined coefficients): z = A cos γt + B sin γt. A particular solution: z = α ω 2 γ 2 cos γt. 73
General solution: y = A sin (ωt + φ 0 ) + α ω 2 cos γt. γ2 forcing by a periodic force, no damping NO RESONANCE superposition of two frequencies
ω/γ rational: periodic motion ω/γ irrational: not periodic 74
Case 2: γ = ω. RESONANCE!!!! y + ω 2 y = α cos ωt General solution, reduced equation: y = C 1 cos ωt + C 2 sin ωt = A sin (ωt + φ 0 ). Form of particular solution (undetermined coefficients): z = A t cos γt + B t sin γt. 75
A particular solution: α 2ω t sin ωt. General solution: y = A sin (ωt + φ 0 ) + α 2ω t sin ωt. Unbounded oscillation This is known as resonance 76
y = A sin (ωt + φ 0 ) + α 2ω t sin ωt. amplitude grows approx. linearly 77
resistance (e.g friction), but no forcing 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 78
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. alpha not zero => roots have a nonzero real part
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, complex roots α 2 4β > 0, two real roots α 2 4β = 0. double (real) root 79
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) 80
or decaying oscillation 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. 81
Underdamped Case: 82
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. decays 83
NOTE: Since α 2 4β < α 2 = α, r 1 and r 2 are both negative and y(t) 0 as t. 84
Case 3: α 2 4β = 0. One real root: (Critically Damped) resonance!!! 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. linear growth defeated by the exponential decay 85
Overdamped and Critically Damped Cases: 86
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. 87
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, 88
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 = a cos γt 89
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). 90
Particular solution of (N) y + α y + β y = a cos γt will have the form: Z(t) = A cos γt + B sin γt. General solution of (N): Note: y(t) = Y c (t) + Z(t) LARGE amplification if gamma is close to omega lim y(t) = Z(t) t 91
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. 92
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 what "survives" for large t 93
Transient solution: Steady-state solution: 94
y = 2e t/4 + e t/2 + 56 85 cos t 48 85 sin t 95
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 cos 2t (C 1 = 2, C 2 = 0) Steady-State solution: Z(t) = 1 5 cos t + 1 10 sin t 96 b/c of damping, the sols of the hom. eq's decay (exponentially fast)
Transient solution. Steady-State solution: 97
higher order linear ODE's with (non-)constant coefficients very similar to the case of 2nd order ODEs Ch 4. Laplace transforms
have one part that is undet coeff and one that is variation of constants at 13:
y = 2e t cos 2t + 1 5 cos t + 1 10 sin t 98