Procedure used to solve equations of the form

Transcription

1 Equations of the form a d 2 y dx 2 + bdy dx + cy = 0 (5) Procedure used to solve equations of the form a d 2 y dx 2 + b dy dx 1. rewrite the given differential equation + cy = 0 (1) a d 2 y dx 2 + b dy + cy = 0 dx (2) (ad 2 + bd + c)y = 0 (3) 2. Substitute m for D and solve the auxillary equation (am 2 + bm + c)y = 0 (4)

2 3 different types of solutions 1. Roots of auxillary equation are REAL and DIFFERENT say m = α and m = β, then the general solution is y = Ae αx + Be βx (6) 2. If the roots of the auxillary equation are real and equal, say m = α twice then, the general solution is y = (Ax + B)e αx (7) 3. if the roots of the auxiliary equation are complex, say m = α ± jβ then the general solution is y = e αx (Ccosβx + Dsinβx) (8)

3 Equations of the form d 2 y dx 2 ± n2 y = 0 1. d 2 y dx 2 +n2 y = 0 m 2 +n 2 = 0 m 2 = n 2 m = ±jn (9) This is a special case of C above α = 0 and β = n.

4 Solution y = e 0x (Ccosβx + Dsinβx) (10) y = (Ccosβx + Dsinβx) (11) d 2 y dx 2 n2 y = 0 m 2 n 2 = 0 m 2 = n 2 m = ±n (12) This is a special case of 1 above. y = Ae αx + Be βx (13)

5 If the particular solution of a differential equation is required then substitute the given boundary conditions to find the unknown constants.

6 Example 1 Find the general solution of 6 d 2 y dx 2 + 5dy dx 4y = 0 (14) and then find the particular solution given that when x=0, y=11 and dy dx =0

7 Solution 1. In D operator form (6D 2 + 5D 4)y = 0 (15) where D = d dx 2. The auxiliary equation is factorising gives 6m 2 + 5m 4 = 0 (16) (3m + 4)(2m 1) = 0 (17) That is m = 4 3 or m = 1 2

8 Since both roots are real and different the general solution is y = Ae 4 3 x + Be 1 2 x (18) When x=0, y=11 then 11 = Ae Be (19) 11 = A + B (20)

9 The other condition we have is that when x=0 dy dx =0 since y = Ae 4 3 x + Be 1 2 x (21) dy dx = 4 3 Ae 4 3 x Be 1 2 x (22)

10 Thus 0 = 4 3 A B (23) 2 0 = 8 3 A + B (24) so B = 8 3A Sub in above A=3 and B=8 11 = A A = 11 3 A (25) y = 3e 4 3 x + 8e 1 2 x (26)

11 Exercise 1 Determine the general solution of 4 d 2 y dt 2 12dy + 9y = 0 (27) dt and then find the particular solution given that when t=0, y=2 and dy dt =4

12 Solution 1. Write in D operator form (4D 2 12D + 9)y = 0 (28) where 2. The Auxiliary equation is D = d dt (29) (4m 2 12m + 9)y = 0 (30) (2m 3)(2m 3) = 0 (31) so m 3 2 twice

13 Since the two roots are real and equal the general solution is y = (At + B)e 3 2 t (32) To find the particular solution when y=2, t=0 2 = (A0 + B)e 3 2 t (33) B=2

14 When dy dt = 4, t = 0 Thus dy dt = (At + B)3 2 e 3 2 t + e 3 2 t (A) (34) 4 = () + B) 3 2 e e (A) (35) So A=1 The particular solution is 4 = 3 2 B + A (36) y = (t + 2)e 3 2 t (37)

15 Exercise 2 Determine the general solution of the DE d 2 y dx 2 + 2dy dx + 5y = 0 (38) and find the particular solution given that when x=0, y=1 and dy dx = 5

16 Solution y = e x (cos2x + 3sin2x) (39)

17 Exercise 3 Determine the general solution of the DE Solution d 2 y 16y = 0 (40) dx 2 y = Ae 4x + Be 4x (41)

18 Exercise 4 Determine the general solution of the DE Solution d 2 y + 49y = 0 (42) dx 2 y = Acos7x + Bsin7x (43)

19 Equations of the Form a d 2 y dx 2 + bdy dx + cy = f (x) (45) The DE a d 2 y dx 2 + b dy + cy = f (x) (44) dx where a,b,c are constants, is a linear, second oreder differential equation with constant coefficients. In order to solve this type of equation a substitution is made.

20 Let y=u+v a d 2 dx 2 (u + v) + b d (u + v) + c(u + v) = f (x) (46) dx (a d 2 u dx 2 + b du dx + cu) + (ad 2 v dx 2 + b dv dx + cv) = f (x) (47)

21 Let v be any solution such that a d 2 v dx 2 + b dv dx + cv = f (x) (48) This means that (a d 2 u dx 2 + b du + cu) = 0 (49) dx We learned how to solve this in the last section

22 The final solution y = u + v (50) The function u is called the Complementary Function(C.F) The function v is called the particular integral(p.i) y = u + v = C.F + P.I (51)

23 Example 1: f(x)= a constant If f(x) is a constant then first try v=k, where k is a constant. Solve d 2 y dx 2 + 4dy dx + 3y = 6 (52)

24 Solution (a d 2 u dx 2 + b du 2 dx + cu) + (ad v dx 2 + b dv + cv) = f (x) (53) dx The complementary function u is obtained as in the previous section (a d 2 u dx 2 + b du + cu) = 0 (54) dx

25 The auxilliary eqn is m 2 + 4m + 3 = 0 (55) (m + 3)(m + 1) = 0 (56) so m=-3 and m=-1

26 2 roots are real and different so the complementary function u = Ae 3x + Be x (57)

27 To get the particular integral we let v=k (a d 2 v dx 2 + b dv dx (a d 2 k dx 2 + b dk dx + cv) = 6 (58) + ck) = 6 (59) (D 2 + 4D + 3)k = 6 (60)

28 D 2 k = 0, D(k) = 0, then 3k = 6, and = 2 (61) Hence the particular integral v=2

29 The general solution is given by y=u+v y = Ae 3x + Be x + 2 (62)

30 Exercise 1: f(x)= a constant If f(x) is a constant then first try v=k, where k is a constant. Solve d 2 y dx 2 5dy + 4y = 20 (63) dx

31 Solution (a d 2 u dx 2 + b du 2 dx + cu) + (ad v dx 2 + b dv + cv) = f (x) (64) dx The complementary function u is obtained as in the previous section

32 The auxilliary eqn is so m=4 and m=1 (a d 2 u dx 2 + b du + cu) = 0 (65) dx m 2 5m + 4 = 0 (66) (m 4)(m 1) = 0 (67)

33 2 roots are real and different so the complementary function u = Ae 4x + Be x (68)

34 To get the particular integral we let v=k (a d 2 v dx 2 + b dv dx (a d 2 k dx 2 + b dk dx + cv) = 20 (69) + ck) = 20 (70) (D 2 5D + 4)k = 20 (71)

35 D 2 k = 0, D(k) = 0, then 4k = 20, and k = 5 (72) Hence the particular integral v=5 The general solution is given by y=u+v y = Ae 4x + Be x + 5 (73)

36 Special Case Find the general solution of d 2 y dx 2 + 4dy dx = 6 (74) and then find the particular solution given that when x=0, y=0 and dy dx = 0.

37 Solution First we find the complementary function u In D operator form the DE is (D 2 + 4D)y = 6 (75) The auxilliary eqn is m 2 + 4m = 0 (76) m(m + 4) = 0 (77)

38 Hence m=0 or -4 We have 2 real roots which are different so u = Ae 0x + Be 4x (78) u = A + Be 4x (79)

39 we must find the particular integral so we assume v=k we obtain which gives us (D 2 + 4D)k = 6 (80) 0 = 6 (81) which is impossible. This occurs because the CF already includes a constant

40 Instead we let v=kx (D 2 + 4D)kx = 6 (82) 4k = 6 (83) k = 3 2 (84) Hence the particular integral v = 3 2 x

41 So the general solution y = u + v = A + Be 4x x (85)

42 Now to find the particular solution When x=0,y=0 and since y = A + Be 4x x (86) 0 = A + Be (87) 0 = A + B (88)

43 dy dx = 0 4Be 4x dy dx = 0, x = 0, 0 = 4B + 3 2, B = 3 8 A = 3 8 (89) (90) (91) Hence the particular solution is y = u + v = e 4x x (92)

44 Exercise 2 Find the general solution of d 2 y dx 2 2dy dx = 4 (93) and then find the particular solution given that when x=0, y=0 and dy dx = 0

45 Solution First we find the complementary function u In D operator form the DE is (D 2 2D)y = 4 (94) The auxilliary eqn is m 2 2m = 0 (95) m(m 2) = 0 (96)

46 Hence m=0 or 2 We have 2 real roots which are different so y = Ae 0x + Be 2x (97) y = A + Be 2x (98)

47 we must find the particular integral so we assume v=k we obtain which gives us (D 2 2D)k = 4 (99) 0 = 4 (100) which is impossible. This occurs because the CF already includes a constant

48 Instead we let v=kx (D 2 2D)kx = 4 (101) 2k = 4 (102) k = 2 (103) Hence the particular integral v = 2x

49 So the general solution y = u + v = A + Be 2x 2x (104)

50 Now to find the particular solution When x=0,y=0 and since y = A + Be 2x 2x (105) 0 = A + Be (0) (106) 0 = A + B (107)

51 dy dx Hence the particular solution is dy dx = 0 + 2Be2x 2 (108) = 0, x = 0, 0 = 2B 2, B = 1 (109) A = 1 (110) y = u + v = 1 + 1e 2x 2x (111)

52 Applications Typical examples of the equation solved above are the equations for the charge q in a series-connected electrical circuit L dq2 dt + R dq dt + 1 C q = E (112) and the force equation for a mass, spring, damped mechanical system m dq2 dt + adq + ks = F (113) dt