Math 211. Lecture #6. Linear Equations. September 9, 2002

Transcription

1 1 Math 211 Lecture #6 Linear Equations September 9, 2002

2 2 Air Resistance

3 2 Air Resistance Acts in the direction opposite to the velocity.

4 2 Air Resistance Acts in the direction opposite to the velocity. Therefore R(x, v) = r(x, v)v where r(x, v) 0.

5 2 Air Resistance Acts in the direction opposite to the velocity. Therefore R(x, v) = r(x, v)v where r(x, v) 0. There are many models. We will look at two different cases.

6 2 Air Resistance Acts in the direction opposite to the velocity. Therefore R(x, v) = r(x, v)v where r(x, v) 0. There are many models. We will look at two different cases. 1. R = rv

7 2 Air Resistance Acts in the direction opposite to the velocity. Therefore R(x, v) = r(x, v)v where r(x, v) 0. There are many models. We will look at two different cases. 1. R = rv, v mg + rv = m.

8 2 Air Resistance Acts in the direction opposite to the velocity. Therefore R(x, v) = r(x, v)v where r(x, v) 0. There are many models. We will look at two different cases. 1. R = rv, v mg + rv = m. 2. R = k v v

9 2 Air Resistance Acts in the direction opposite to the velocity. Therefore R(x, v) = r(x, v)v where r(x, v) 0. There are many models. We will look at two different cases. 1. R = rv, v mg + rv = m. 2. R = k v v, v = mg + k v v m.

10 3 Solving for x(t) Resistance

11 3 Solving for x(t) Integrating x = v(t) is sometimes hard. Resistance

12 3 Solving for x(t) Integrating x = v(t) is sometimes hard. Use the trick (see Exercise 2.3.7): Resistance

13 3 Solving for x(t) Integrating x = v(t) is sometimes hard. Use the trick (see Exercise 2.3.7): a = dv dt Resistance

14 3 Solving for x(t) Integrating x = v(t) is sometimes hard. Use the trick (see Exercise 2.3.7): a = dv dt = dv dx dx dt Resistance

15 3 Solving for x(t) Integrating x = v(t) is sometimes hard. Use the trick (see Exercise 2.3.7): a = dv dt = dv dx dx dt = dv dx v Resistance

16 3 Solving for x(t) Integrating x = v(t) is sometimes hard. Use the trick (see Exercise 2.3.7): a = dv dt = dv dx dx dt = dv dx v The equation is usually separable. v dv dx = a = F m Resistance

17 4 Problem A ball is projected from the surface of the earth with velocity v 0. How high does it go? Resistance

18 4 Problem A ball is projected from the surface of the earth with velocity v 0. How high does it go? At t =0, we have x(0)=0and v(0) = v 0. Resistance

19 4 Problem A ball is projected from the surface of the earth with velocity v 0. How high does it go? At t =0, we have x(0)=0and v(0) = v 0. At the top we have t = T, x(t )=x max, and v(t )=0. Resistance

20 4 Problem A ball is projected from the surface of the earth with velocity v 0. How high does it go? At t =0, we have x(0)=0and v(0) = v 0. At the top we have t = T, x(t )=x max, and v(t )=0. We have Resistance

21 4 Problem A ball is projected from the surface of the earth with velocity v 0. How high does it go? At t =0, we have x(0)=0and v(0) = v 0. At the top we have t = T, x(t )=x max, and v(t )=0. We have v dx dv = a = F m = mg + R, so m 0 vdv xmax mg R = dx 0 m. v 0 Resistance

22 5 Cases Problem Resistance

23 5 Cases 1. R =0. Problem Resistance

24 5 Cases 1. R =0. x max = v2 0 2g. Problem Resistance

25 5 Cases 1. R =0. x max = v2 0 2g. 2. R = rv. Problem Resistance

26 5 Cases 1. R =0. x max = v2 0 2g. 2. R = rv. x max = m r [ v 0 mg r ln ( 1+ rv )] 0. mg Problem Resistance

27 5 Cases 1. R =0. x max = v2 0 2g. 2. R = rv. x max = m r [ v 0 mg r ln ( 1+ rv )] 0. mg 3. R = k v v. Problem Resistance

28 5 Cases 1. R =0. x max = v2 0 2g. 2. R = rv. x max = m r [ v 0 mg r ln ( 1+ rv )] 0. mg 3. R = k v v. x max = m 2k ln ( 1+ kv2 0 mg ). Problem Resistance

29 6 Linear Equations

30 6 Linear Equations A linear equation is one of the form x = a(t)x + f(t).

31 6 Linear Equations A linear equation is one of the form x = a(t)x + f(t). Example: x = tan(t)x + 3 sin 2 (t)

32 6 Linear Equations A linear equation is one of the form x = a(t)x + f(t). Example: x = tan(t)x + 3 sin 2 (t) The unknown function x and its derivative must appear linearly.

33 6 Linear Equations A linear equation is one of the form x = a(t)x + f(t). Example: x = tan(t)x + 3 sin 2 (t) The unknown function x and its derivative must appear linearly. The equation is homogeneous if f =0

34 6 Linear Equations A linear equation is one of the form x = a(t)x + f(t). Example: x = tan(t)x + 3 sin 2 (t) The unknown function x and its derivative must appear linearly. The equation is homogeneous if f =0 x = a(t)x

35 6 Linear Equations A linear equation is one of the form x = a(t)x + f(t). Example: x = tan(t)x + 3 sin 2 (t) The unknown function x and its derivative must appear linearly. The equation is homogeneous if f =0 x = a(t)x, e.g. x = tan(t)x

36 6 Linear Equations A linear equation is one of the form x = a(t)x + f(t). Example: x = tan(t)x + 3 sin 2 (t) The unknown function x and its derivative must appear linearly. The equation is homogeneous if f =0 x = a(t)x, e.g. x = tan(t)x The equation is inhomogeneous if f 0

37 7 Homogenous Linear Equations

38 7 Homogenous Linear Equations Homogeneous linear equations are separable. dx dt = a(t)x

39 7 Homogenous Linear Equations Homogeneous linear equations are separable. dx dt = a(t)x x(t) =Ae a(t) dt

40 7 Homogenous Linear Equations Homogeneous linear equations are separable. Example: x = tan(t)x. dx dt = a(t)x x(t) =Ae a(t) dt

41 7 Homogenous Linear Equations Homogeneous linear equations are separable. Example: x = tan(t)x. dx dt = a(t)x x(t) =Ae a(t) dt x(t) =Ae sec2 t

42 8 Example: x = tan(t)x + 3 sin 2 (t)

43 8 Example: x = tan(t)x + 3 sin 2 (t) Rewrite as x tan(t)x = 3 sin 2 (t)

44 8 Example: x = tan(t)x + 3 sin 2 (t) Rewrite as x tan(t)x = 3 sin 2 (t) Multiply by cos t. cos(t)x sin(t)x = 3 sin 2 (t) cos(t)

45 8 Example: x = tan(t)x + 3 sin 2 (t) Rewrite as x tan(t)x = 3 sin 2 (t) Multiply by cos t. cos(t)x sin(t)x = 3 sin 2 (t) cos(t) The left hand side is the derivative of cos(t)x.

46 8 Example: x = tan(t)x + 3 sin 2 (t) Rewrite as x tan(t)x = 3 sin 2 (t) Multiply by cos t. cos(t)x sin(t)x = 3 sin 2 (t) cos(t) The left hand side is the derivative of cos(t)x. So [cos(t)x] = 3 sin 2 (t) cos(t)

47 9 Integrate Solution pt. 1

48 9 Integrate cos(t)x(t) =3 sin 2 (t) cos(t) dt = sin 3 (t)+c Solution pt. 1

49 9 Integrate cos(t)x(t) =3 sin 2 (t) cos(t) dt = sin 3 (t)+c Solve for x x(t) = sin3 (t)+c cos(t). Solution pt. 1

50 9 Integrate cos(t)x(t) =3 sin 2 (t) cos(t) dt = sin 3 (t)+c Solve for x x(t) = sin3 (t)+c cos(t) How did we do that?. Solution pt. 1

51 9 Integrate cos(t)x(t) =3 sin 2 (t) cos(t) dt = sin 3 (t)+c Solve for x x(t) = sin3 (t)+c. cos(t) How did we do that? Can we do it in general? Solution pt. 1

52 10 The Key Step for x = ax + f Solution pt. 1

53 10 The Key Step for x = ax + f Rewrite as x ax = f. Solution pt. 1

54 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that Solution pt. 1

55 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that u[x ax] =[ux] Solution pt. 1

56 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that u[x ax] =[ux] ux aux Solution pt. 1

57 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that u[x ax] =[ux] ux aux = ux + u x Solution pt. 1

58 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that u[x ax] =[ux] ux aux = ux + u x True if u = au. Solution pt. 1

59 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that u[x ax] =[ux] ux aux = ux + u x True if u = au. Linear, homogeneous Solution pt. 1

60 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that u[x ax] =[ux] ux aux = ux + u x True if u = au. Linear, homogeneous u(t) =e a(t) dt is one solution. Solution pt. 1

61 10 The Key Step for x = ax + f Rewrite as x ax = f. Multiply by a function u(t) so that u[x ax] =[ux] ux aux = ux + u x True if u = au. Linear, homogeneous u(t) =e a(t) dt is one solution. u is called an integrating factor. Solution pt. 1

62 11 Solving the Linear Equation x = a(t)x + f(t)

63 11 Four step process: Solving the Linear Equation x = a(t)x + f(t)

64 11 Solving the Linear Equation x = a(t)x + f(t) Four step process: 1. Rewrite as x ax = f.

65 11 Solving the Linear Equation x = a(t)x + f(t) Four step process: 1. Rewrite as x ax = f. 2. Multiply by the integrating factor

66 11 Four step process: Solving the Linear Equation x = a(t)x + f(t) 1. Rewrite as x ax = f. 2. Multiply by the integrating factor u(t) =e a(t) dt.

67 11 Four step process: Solving the Linear Equation x = a(t)x + f(t) 1. Rewrite as x ax = f. 2. Multiply by the integrating factor u(t) =e a(t) dt. Equation becomes [ux] = ux aux = uf.

68 11 Four step process: Solving the Linear Equation x = a(t)x + f(t) 1. Rewrite as x ax = f. 2. Multiply by the integrating factor u(t) =e a(t) dt. Equation becomes [ux] = ux aux = uf. 3. Integrate: u(t)x(t) = u(t)f(t) dt + C.

69 11 Four step process: Solving the Linear Equation x = a(t)x + f(t) 1. Rewrite as x ax = f. 2. Multiply by the integrating factor u(t) =e a(t) dt. Equation becomes [ux] = ux aux = uf. 3. Integrate: u(t)x(t) = u(t)f(t) dt + C. 4. Solve for x(t).

70 12 Examples Solution method

71 12 Examples x = 4x +8, x(0)=0. Solution method

72 12 Examples x = 4x +8, x =2tx + e t2, x(0)=0. x(0)=1. Solution method

73 12 Examples x = 4x +8, x =2tx + e t2, x(0)=0. x(0)=1. y =3y t, y(0)=2. Solution method

74 12 Examples x = 4x +8, x =2tx + e t2, x(0)=0. x(0)=1. y =3y t, y(0)=2. z =(z + 1) cos t, z(π) = 1. Solution method