Advanced Eng. Mathematics

Transcription

1 Koya University Faculty of Engineering Petroleum Engineering Department Advanced Eng. Mathematics Lecture 6 Prepared by: Haval Hawez haval.hawez@koyauniversity.org 1

2 Second Order Linear Ordinary Differential Equations

3 Second-Order Linear ODEs: ODEs might be separated into two large classes: Linear ODEs and non-linear ODEs. d 2 y dy + x + 2y = sinx dx2 dx (Linear Second Order DE) d 2 y dy dx2 y + x + 2y = sinx dx (nonlinear Second Order DE) Second Order Linear ODES 3 3

4 Second-Order Linear ODEs: None-linear ODEs of second and higher order are generally difficult to solve. Linear Second Order ODEs are much simpler because various properties of their solutions can be characterized in a general way. Second Order Linear ODES 4 4

5 Second-Order Linear ODEs: Linear ODEs of the second order are the most important ones because of their applications in mechanical and electrical engineering. And their theory is typical of that of all linear ODEs, but the formulas are simpler than for higher order equations. Second Order Linear ODES 5 5

6 Homogenous Linear ODEs of Second Order A second-order ODE is called linear if can be written y + p x y + q x y = r(x) (1) If r x = 0 (that is, r x = 0 for all x considered; read r(x) is identically zero ), then (1) reduces to y + p x y + q x y = 0 (2) And is called homogeneous. If r(x) 0, then (1) is called nonhomogeneous. Second Order Linear ODES 6 6

7 Homogenous Linear ODEs od Second Order For instance, a nonhomogeneous linear ODE is y + 25y = e x cos x And a homogeneous linear ODE is xy + y + xy = 0, in standard form y + 1 x y + y = 0 Second Order Linear ODES 7 7

8 Example 1 Homogenous Linear ODEs: Solve y + y = 0 when y = cosx For all x. We verify this by differentiation and substitution. We obtain cos x = cos x; hence y + y = cos x + cos x y + y = cos x + cos x = 0. Second Order Linear ODES 8 8

9 Example 1 Homogenous Linear ODEs: Solve y +y = 0 Similarly for y = sin x (verify!). We can go an important step further. We multiply cos x by any constant, for instance, 4.7. and sin x by, say, -2. And take the sum of the results, claiming that it is a solution. Indeed, differentiation and substitution gives Second Order Linear ODES 9 9

10 Example 1 Homogenous Linear ODEs: Solve y + y = 0 Similarly for y = sin x (verify!). y + y = 4.7 cos x 2 sin x cos x 2 sin x y + y = 4.7 cos x + 2 sin x cos x 2 sin x = 0. In this example we have obtained from y 1 = cos x and y 2 (= sin x) Second Order Linear ODES 10 10

11 Example 1 Homogenous Linear ODEs: Solve y + y = 0 Similarly for y = sin x (verify!). y = c 1 y 1 + c 2 y 2 a function of the form (3) (c 1, c 2 arbitrary constants). This is called a linear combination of y 1 and y 2. In terms of this concept we can now formulate the result suggested by our example, often called the superposition principle or linearity principle. Second Order Linear ODES 11 11

12 Theorem 1 Fundamental Theorem for the Homogeneous Linear ODE (2) Proof Let y 1 and y 2 be solutions of (2) on I. then by substituting y = c 1 y 1 + c 2 y 2 and its derivatives into (2), and using the familiar rule c 1 y 1 + c 2 y 2 = c 1 y 1 + c 2 y 2, etc., we get y + py + qy = c 1 y 1 + c 2 y 2 + p c 1 y 1 + c 2 y 2 + q(c 1 y 1 + c 2 y 2 ) = c 1 y 1 + c 2 y 2 + p c 1 y 1 + c 2 y 2 + q(c 1 y 1 + c 2 y 2 ) = c 1 y 1 + py 1 + qy 1 + c 2 y 2 + py 2 + qy 2 = 0, Since in the last time, ( ) = 0 because y 1 and y 2 are solutions, by assumption. This shows that y is a solution of (2) on I. Second Order Linear ODES 12 12

13 Homogenous Linear ODEs with CC: We shall now consider second-order homogeneous linear ODEs whose coefficients a and b are constant, y + ay + by = 0. (1) These equations have important applications, especially in connection with mechanical and vertical vibrations. y + ky = 0 Homogenous Linear ODEs with CC 13 13

14 Homogenous Linear ODEs with CC: Is an exponential function y = ce kx. This gives us the idea to try as a solution of (1) the function y = e λx (2) Substituting (2) and its derivatives y = λe λx and y = λ 2 e ℵx Into our equation (1), we obtain λ 2 + aλ + b e λx = 0. Hence if λ is a solution of the important characteristic equation (or auxiliary equation) λ 2 + aλ + b = 0 (3) Homogenous Linear ODEs with CC 14 14

15 Homogenous Linear ODEs with CC: Then the exponential function (2) is a solution of the ODE (1). Now from elementary algebra we recall that the roots of this quadratic equation (3) are λ1 = 1 2 a + a2 4b, λ2 = 1 2 a a2 4b (4) (3) and (4) will be basic because our derivation shows that the functions y 1 = e λ1x and y 2 = e λ2x (5) Homogenous Linear ODEs with CC 15 15

16 Homogenous Linear ODEs with CC: Are solutions of (1). Verify this by substituting (5) into (1). From algebra we further know that the quadratic equation (3) may have three kinds of roots, depending on the sign of the discriminanta 2 4b, namely, (Case I) Two real roots if a 2 4b > 0, (Case II) A real double root if a 2 4b = 0, (Case III) Complex conjugate roots if a 2 4b < 0. Homogenous Linear ODEs with CC 16 16

17 Case I. Two Distinct Real Roots λ1 and λ2 In this case, a basis of solutions of (1) on any interval is y 1 = e λ1x and y 2 = e λ2x Because y 1 and y 2 are defined (and real) for all x and their quotient is not constant. The corresponding general solution is y = c 1 e λ1x + c 2 e λ2x. (6) Homogenous Linear ODEs with CC 17 17

18 Example 1: General Solution Solve y y = 0. The characteristic equation is λ 2 1 = 0. its roots are λ 1 = 1 and λ 2 = 1. Hence a basis of solutions is e x and e x and gives the same general solution as before. y = c 1 e λ1x + c 2 e λ2x. Homogenous Linear ODEs with CC 18 18

19 Example 2: Initial Value Problem Solve the initial value problem y + y 2y = 0, y 0 = 4, y 0 = 5. Solution. Step1. General solution. The characteristic equation is λ 2 + λ 2 = 0. Its roots are λ 1 = = 1 and λ 2 2 = 1 2 So that we obtain the general solution y = c 1 e x + c 2 e 2x. 1 9 = 2 Homogenous Linear ODEs with CC 19 19

20 Example 2: Initial Value Problem Solve the initial value problem y + y 2y = 0, y 0 = 4, y 0 = 5. Step 2. Particular solution. Since y x = c 1 e x 2 c 2 e 2x, we obtain from the general solution and the initial conditions y 0 = c 1 + c 2 = 4. y 0 = c 1 2c 2 = 5. Hence c 1 = 1 and c 2 = 3. This gives the answer y = e x + 3e 2x. Homogenous Linear ODEs with CC 20 20

21 Case II. Real Double Root λ = a 2 If the discriminant a 2 4b is zero, we see directly from equation below (4) λ1 = 1 2 a + a2 4b λ2 = 1 2 a a2 4b that we get only one root, λ = λ 1 = λ 2 = a 2, hence only one solution, y 1 = e a 2 x. Case II & Case III 21

22 Case II. Real Double Root λ = a 2 To obtain a second independent solution y 2 (needed for a basis), we use the method of reduction of order discussed in the last section, setting y 2 = uy 1. Substituting this and its derivatives y 2 = u y 1 + uy 1 and y 2 into (1), we first have u y 1 + 2u y 1 + uy 1 + a u y 1 + uy 1 + buy 1 = 0. Case II & Case III 22

23 Case II. Real Double Root λ = a 2 Collecting terms in u, u, and u, as in the last section, we obtain u y 1 + u 2y 1 + ay 1 + u y 1 + ay 1 + by 1 = 0. The expression in the last parentheses is zero, since y 1 is a solution of (1). The expression in the first parentheses is zero, too, since 2y 1 = ae ax 2 = ay 1. y = (c 1 + c 2 x)e λx Case II & Case III 23

24 Example 3: General Solution The characteristic equation of the ODE y + 6y + 9y = 0 Solution: λ 2 + 6λ + 9 = 0. a = 6, b = 9 a 2 4b = = = 0 It is Case II λ = λ 1 = λ 2 = a 2 = 6 2 = 3 Case II & Case III 24

25 Example 3: General Solution The characteristic equation of the ODE y + 6y + 9y = 0 Solution: It has the double root λ = 3. Hence a basis is e 3x and xe 3x. The corresponding general solution is y = (c 1 + c 2 x)e 3x. Case II & Case III 25

26 Example 4: Initial Value Problem Case II Solve the initial value problem y + y y = 0, y 0 = 3, y 0 = 3.5. Solution. The characteristic equation is λ 2 + λ a = 1, b = a 2 4b = = 1 1 = 0 It is Case II λ = λ 1 = λ 2 = a 2 = 1 2 = 0. 5 Case II & Case III 26

27 Example 4: Initial Value Problem Case II Solve the initial value problem y + y y = 0, y 0 = 3, y 0 = 3.5. Solution: This gives the general solution y = (c 1 + c 2 x)e λx y = (c 1 + c 2 x)e 0.5x Case II & Case III 27

28 Example 4: Initial Value Problem Case II Solve the initial value problem y + y y = 0, y 0 = 3, y 0 = 3.5. Solution: We need its derivative y = c 2 e 0.5x 0. 5 c 1 + c 2 x e 0.5x. From this and the initial conditions we obtain y 0 = c 1 = 3, y 0 = c 2 0.5c 1 = 3.5 hence c 2 = 2. The particular solution of the initial value problem is y = 3 2x e 0.5x Case II & Case III 28

29 Case III. Complex Roots 1 a + iω and 1 a iω 2 2 This case occurs if the discriminant a 2 4b characteristic equation (3) is negative. of the In this case, the roots of (3) and thus the solutions of the ODE (1) come at first out complex. However, we show that from them we can obtain a basis of real solutions y 1 = e ax/2 cos ωx, (8) y 2 = e ax/2 sin ωx, ω > 0 Case II & Case III 29

30 Case III. Complex Roots 1 a + iω and 1 a iω 2 2 Where ω 2 = b 1 4 a2 It can be verified by substitution that these are solutions in the present case. We shall derive them systematically after the two examples by using the complex exponential function. They form a basis on any interval since their quotient cot ωx is not constant. Case II & Case III 30

31 Case III. Complex Roots 1 a + iω and 1 a iω 2 2 Hence a real general solution in Case III is y = e ax/2 (A cos ωx + B sin ωx) (A, B arbitrary constant) (9) Case II & Case III 31

32 Example 5: Complex Roots Solve the initial value problem y + 0.4y y = 0, y 0 = 0, y 0 = 3. Solution. Step 1. General solution. The characteristic equation is λ λ = 0. a = 0. 4, b = a 2 4b = = 36 Case II & Case III 32

33 Example 5: Complex Roots Solve the initial value problem y + 0.4y y = 0, y 0 = 0, y 0 = 3. Solution. Step 1. General solution. ω 2 = b 1 4 a2 ω 2 = = = 9 Hence ω = 3, and a general solution is y = e 0.2x (A cos 3x + B sin 3x) Case II & Case III 33

34 Example 5: Complex Roots Solve the initial value problem y + 0.4y y = 0, y 0 = 0, y 0 = 3. Step 2. Particular solution. The first initial condition gives y 0 = A = 0. The remaining expression is y = Be 0.2x sin 3x. Case II & Case III 34

35 Example 5: Complex Roots Solve the initial value problem y + 0.4y y = 0, y 0 = 0, y 0 = 3. Step 2. Particular solution. We need the derivative (chain rule!) y = B 0.2 e 0.2x sin 3x + 3e 0.2x cos 3x. From this and the second initial condition we obtain y 0 = 3B = 3. Hence B=1. Our solution is y = e 0.2x sin 3x. Case II & Case III 35