Differential Equations Review

Transcription

1 P. R. Nelson diff eq prn.tex Winter 2010 p. 1/20 Differential Equations Review Phyllis R. Nelson Professor, Department of Electrical and Computer Engineering California State Polytechnic University, Pomona

2 Examples Classical mechanics: Newton s law of cooling: F = m d2 x dt 2 Wave equation: dt dt = k(t T 0) 2 y t 2 = v2 2 y x 2 P. R. Nelson diff eq prn.tex Winter 2010 p. 2/20

3 Types of variables Independent variables: Dependent variables: Parameters: dt dt = k(t T 0) 2 y t 2 = v2 2 y x 2

4 P. R. Nelson diff eq prn.tex Winter 2010 p. 3/20 Types of variables Independent variables: Dependent variables: Parameters: dt dt = k(t T 0) dt dt = k(t T 0) 2 y t 2 = v2 2 y x 2 2 y t 2 = v2 2 y x 2

5 Ordinary vs. partial Ordinary:

6 Ordinary vs. partial Ordinary: one independent variable

7 Ordinary vs. partial Ordinary: one independent variable Partial:

8 Ordinary vs. partial Ordinary: one independent variable Partial: more than one independent variable

9 P. R. Nelson diff eq prn.tex Winter 2010 p. 4/20 Ordinary vs. partial Ordinary: one independent variable Partial: more than one independent variable dt dt = k(t T 0) 2 y t 2 = v2 2 y x 2

10 P. R. Nelson diff eq prn.tex Winter 2010 p. 5/20 Order The order of the highest derivative dy dx e y d2 y dx 2 + y + ax = b3 ( ) 3 dy = 0 dx 2 y t 2 = v2 2 y x 2

11 P. R. Nelson diff eq prn.tex Winter 2010 p. 6/20 Degree The power to which the highest-order derivative is raised dy dx e y d2 y dx 2 + y + ax = b3 ( ) 3 dy = 0 dx 2 y t 2 = v2 2 y x 2

12 P. R. Nelson diff eq prn.tex Winter 2010 p. 7/20 Constraints Initial-value problems: the constraints are specified at a single value of the independent variable d 2 z dt = mg z(t = 0) = 0 dz 2 dt = 0 t=0

13 P. R. Nelson diff eq prn.tex Winter 2010 p. 8/20 Constraints (2) Boundary-value problems: the constraints are specified at multiple values of the independent variable(s) d 2 y + 5y = 0 y(x = 0) = 0 y(x = 8) = 0 dx2

14 P. R. Nelson diff eq prn.tex Winter 2010 p. 9/20 OLDECC Ordinary Linear Differential Equations with Constant Coefficients Useful abbreviations: dy dx y d 2 y dx 2 y d n y dx n y(n) To solve, guess the solution y = e γx

15 Example 1 y + 2y + 10y = 0 (1) Step 1: Guess y = e γx and try it out in (1).

16 Example 1 y + 2y + 10y = 0 (3) Step 1: Guess y = e γx and try it out in (1). y = γe γx y = γ 2 e γx

17 Example 1 y + 2y + 10y = 0 (5) Step 1: Guess y = e γx and try it out in (1). y = γe γx y = γ 2 e γx γ 2 e γx + 2γe γx + 10e γx = ( γ 2 + 2γ + 10 ) e γx = 0

18 P. R. Nelson diff eq prn.tex Winter 2010 p. 10/20 Example 1 y + 2y + 10y = 0 (7) Step 1: Guess y = e γx and try it out in (1). y = γe γx y = γ 2 e γx γ 2 e γx + 2γe γx + 10e γx = ( γ 2 + 2γ + 10 ) e γx = 0 γ 2 + 2γ + 10 = 0 (8)

19 Example 1 (2) Equation (2) is called the auxiliary equation. Step 2: Solve the auxiliary equation. Reminder: roots of a quadratic equation aγ 2 + bγ + c = 0 γ = b ± b 2 4ac 2a (9) a = 1 b = 2 c = 10 Reminder: 1 = i γ = 1 ± 9

20 P. R. Nelson diff eq prn.tex Winter 2010 p. 11/20 Example 1 (2) Equation (2) is called the auxiliary equation. Step 2: Solve the auxiliary equation. Reminder: roots of a quadratic equation aγ 2 + bγ + c = 0 γ = b ± b 2 4ac 2a (10) a = 1 b = 2 c = 10 Reminder: 1 = i γ = 1 ± 9 γ = 1 ± 3i

21 Example 1 (3) So y can be e ( 1+3i)x = e x e 3ix or e ( 1 3i)x = e x e 3ix

22 Example 1 (3) So y can be e ( 1+3i)x = e x e 3ix or e ( 1 3i)x = e x e 3ix Step 3: Use a linear combination of possible solutions as the general solution.

23 Example 1 (3) So y can be e ( 1+3i)x = e x e 3ix or e ( 1 3i)x = e x e 3ix Step 3: Use a linear combination of possible solutions as the general solution. y = e x ( C 1 e 3ix + C 2 e 3ix) (13)

24 P. R. Nelson diff eq prn.tex Winter 2010 p. 12/20 Example 1 (3) So y can be e ( 1+3i)x = e x e 3ix or e ( 1 3i)x = e x e 3ix Step 3: Use a linear combination of possible solutions as the general solution. y = e x ( C 1 e 3ix + C 2 e 3ix) (14) We need a less abstract form. It s hard to picture the graph of y = e i3x and use it for physical reasoning.

25 Example 1 (4) Reminders: (Euler s formula) and e iθ = cosθ + i sin θ cos( θ) = cosθ sin( θ) = sin θ y = e x ( C 1 e 3ix + C 2 e 3ix)

26 P. R. Nelson diff eq prn.tex Winter 2010 p. 13/20 Example 1 (4) Reminders: (Euler s formula) and e iθ = cosθ + i sin θ cos( θ) = cosθ sin( θ) = sin θ y = e x ( C 1 e 3ix + C 2 e 3ix) = e x (A cos 3x + B sin 3x)

27 P. R. Nelson diff eq prn.tex Winter 2010 p. 14/20 Example 1 (5) y = e x (A cos 3x + B sin 3x) (15) This is the general solution. Since the original equation y + 2y + 10y = 0 is second order, there are two parameters (A and B) that must be found from the initial or boundary conditions.

28 Example 1 (6) Given initial conditions y(0) = 0 y (0) = K Substitute into y = e x (A cos 3x + B sin 3x)

29 P. R. Nelson diff eq prn.tex Winter 2010 p. 15/20 Example 1 (6) Given initial conditions y(0) = 0 y (0) = K Substitute into y = e x (A cos 3x + B sin 3x) y(0) = e 0 [A cos 3(0) + B sin 3(0)] = 1 [A(1) + B(0)] = A = 0

30 Example 1 (7) y (x) = d dx [ e x (B sin 3x) ]

31 Example 1 (7) y (x) = d dx [ e x (B sin 3x) ] = e x (B sin 3x) + e x (3B cos 3x) = Be x (3 cos 3x sin 3x)

32 P. R. Nelson diff eq prn.tex Winter 2010 p. 16/20 Example 1 (7) y (x) = d dx [ e x (B sin 3x) ] = e x (B sin 3x) + e x (3B cos 3x) = Be x (3 cos 3x sin 3x) y (0) = B(1) [3 cos 3(0) sin 3(0)] = 3B = K B = K 3

33 P. R. Nelson diff eq prn.tex Winter 2010 p. 17/20 Example 1 (8) y = ( K 3 ) e x sin 3x Script to display a graph in octave: k=1; x = linspace(0,5,100); y = k./3.*exp(-1.*x).* sin(3.*x); plot(x,y)

34 Example 2 y + 4y = 0 (16)

35 Example 2 Guess y = e γx y + 4y = 0 (18)

36 Example 2 y + 4y = 0 (20) Guess y = e γx γ = ±2i

37 Example 2 y + 4y = 0 (22) Guess y = e γx γ = ±2i Let y be a linear combination of possible solutions.

38 Example 2 y + 4y = 0 (24) Guess y = e γx γ = ±2i Let y be a linear combination of possible solutions. y = C 1 e 2ix + C 2 e 2ix

39 Example 2 y + 4y = 0 (26) Guess y = e γx γ = ±2i Let y be a linear combination of possible solutions. y = C 1 e 2ix + C 2 e 2ix Simplify.

40 P. R. Nelson diff eq prn.tex Winter 2010 p. 18/20 Example 2 y + 4y = 0 (28) Guess y = e γx γ = ±2i Let y be a linear combination of possible solutions. y = C 1 e 2ix + C 2 e 2ix Simplify. y = A cos 2x + B sin 2x (29)

41 Example 2 (2) Given boundary conditions y(0) = 0 y( π 4 ) = 2

42 Example 2 (2) Given boundary conditions y(0) = 0 y( π 4 ) = 2 Substitute the general solution.

43 Example 2 (2) Given boundary conditions y(0) = 0 y( π 4 ) = 2 Substitute the general solution. y(0) = A = 0

44 Example 2 (2) Given boundary conditions y(0) = 0 y( π 4 ) = 2 Substitute the general solution. y(0) = A = 0 y( π 4 ) = B sin 2 ( π 4) = B = 2

45 P. R. Nelson diff eq prn.tex Winter 2010 p. 19/20 Example 2 (2) Given boundary conditions y(0) = 0 y( π 4 ) = 2 Substitute the general solution. y(0) = A = 0 y( π 4 ) = B sin 2 ( π 4) = B = 2 y = 2 sinx

46 P. R. Nelson diff eq prn.tex Winter 2010 p. 20/20 Example 2 (3) Here is a script to display the graph of y(x) in octave: x = linspace(0,pi/4,100); y = 2.* sin(2.*x); plot(x,y)