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Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA.

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Presentation on theme: "Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA."— Presentation transcript:

1 Homogeneous Linear Differential Equations with Constant Coefficients A Lecture in ENGIANA

2 Auxillary Equation Consider a second order equation ay’’ + by’ + cy = 0 where a, b, and c are constants. If we try to find a solution of the form y = e mx, then after substitution of y’ = me mx and y’’ = m 2 e mx, the equation becomes am 2 e mx + bme mx + ce mx = 0

3 Auxillary Equation Solving am 2 e mx + bme mx + ce mx = 0, e mx (am 2 + bm + c) = 0 The quantity in parenthesis, a quadratic equation, is called the auxiliary equation. This means that to find the solution y (see previous slide), we must solve for m.

4 Auxillary Equation There are three possible cases: m 1  m 2 ; distinct real roots m 1 = m 2 ; repeated real roots m 1  m 2 ; conjugate complex roots

5 Case 1: Distinct Real Roots For this case, we have And hence, Or

6 Example Find the general solution of (D 2 + D – 6) y = 0

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8 Case 2: Real Repeated Roots Having two real, repeated roots means Now, one solution is

9 Case 2: Real Repeated Roots Recall that a 2 (x)y’’ + a 1 (x)y’ + a 0 (x)y = 0 can be written as y” + P(x)y’ + Q(x)y = 0 where P(x) = a 1 (x)/a 2 (x) Q(x) = a 0 (x)/a 2 (x)

10 Case 2: Real Repeated Roots In our case, the coefficients are constants: ay’’ + by’ + cy = 0 Thus, y” + Py’ + Qy = 0 where P = b/a Q = c/a

11 Case 2: Real Repeated Roots Recall also that another solution y 2 is

12 Case 2: Real Repeated Roots Hence,

13 Case 2: Real Repeated Roots The general solution is then

14 Example Find the general solution of y’’ + 8y’ + 16y = 0

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16 Case 3: Conjugate Complex Roots If m 1 and m 2 are complex, then we have m 1 =  + i  m 2 =  - i  where  and  are real and positive Hence, we can write y = C 1 e (  + i  )x + C 2 e (  - i  )x

17 Case 3: Conjugate Complex Roots However, in practice we prefer to work with real functions instead of complex exponentials. To this end, we use Euler’s formula: e i  = cos  + isin  where  is any real number

18 Case 3: Conjugate Complex Roots Thus, we have e i  x = cos  x + isin  x e - i  x = cos  x - isin  x Note that e i  x + e - i  x = 2cos  x & e i  x – e - i  x = 2isin  x

19 Case 3: Conjugate Complex Roots Our solution is then y = C 1 e (  +i  )x + C 2 e (  -i  )x If we let C 1 = 1 and C 2 = 1: y 1 = e (  +i  )x + e (  -i  )x y 1 = e  x (e i  x + e -i  x ) y 1 = e  x (2cos  x) y 1 = 2e  x cos  x

20 Case 3: Conjugate Complex Roots If we let C 1 = 1 and C 2 = -1: y 2 = e (  +i  )x - e (  -i  )x y 2 = e  x (e i  x - e -i  x ) y 2 = e  x (2isin  x) y 2 = 2ie  x sin  x

21 Case 3: Conjugate Complex Roots Thus, the solution to y = C 1 e (  + i  )x + C 2 e (  - i  )x is y = c 1 y 1 + c 2 y 2 y = c 1 (e  x cos  x) + c 2 (e  x sin  x) or y = e  x (c 1 cos  x + c 2 sin  x)

22 Example Find the general solution of (D 2 – 4D + 7) y = 0

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24 Higher-Order (n>2) Equations: Distinct Roots Consider the case where the auxiliary equation has distinct roots. Say we are given f(D)y = 0. Then one possible solution is e mx, f(D)e mx = 0, if the auxiliary equation is f(m) = 0

25 Higher-Order (n>2) Equations: Distinct Real Roots In other words, if the distinct roots of the auxiliary equation are m 1, m 2, …, m n, then the corresponding solutions are exp(m 1 x), exp(m 2 x), …, exp(m n x). The general solution is

26 Example Find the general solution of (D 3 + 6D D + 6) y = 0

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28 Higher-Order (n>2) Equations: Repeated Real Roots Consider the case where the auxiliary equation has repeated roots. Say we are given f(D)y = 0. If there are several identically repeated roots m 1 = m 2 = … = m n = b, then this means (D - b) n y = 0

29 Higher-Order (n>2) Equations: Repeated Roots If we let y = x k e bx [k = 0, 1, 2, …, (n-1)] Then, (D – b) n y = (D – b) n [x k e bx ] But (D – b) n [x k e bx ] = e bx D n [x k ] = e bx (0) Thus, (D – b) n y = (D – b) n [x k e bx ] = 0

30 Higher-Order (n>2) Equations: Repeated Roots The functions y k = x k e bx [e.g., e 7x, xe 7x, x 2 e 7x, etc.], where k = 0, 1, 2, …, (n – 1) are linearly independent because, aside from the common factor e bx, they contain only the respective powers x 0, x 1, x 2, …, x n-1. The general solution is thus y = c 1 e bx + c 2 xe bx + … + c n x n-1 e bx

31 Example Find the general solution of (D 4 + 6D 3 + 9D 2 ) y = 0

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33 Higher-Order (n>2) Equations: Repeated Imaginary Roots Repeated imaginary roots lead to solutions analogous to those brought in by repeated real roots. For instance, if the conjugate pair m = a  bi occur three times, the corresponding general solution is y= (c 1 + c 2 x + c 3 x 2 ) e ax cosbx + (c 4 + c 5 x + c 6 x 2 ) e ax sinbx

34 Example Find the general solution of (D D ) y = 0

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36 Exercises Find the solution required: 1)(D 2 – 2D – 3)y = 0 y(0)=0; y’(0)=-4 2)(D 3 – 4D)y = 0 y(0)=0; y’(0)=0; y’’(0)=2 3)(D 4 + 2D D 2 )y = 0 4)(D 6 + 9D D )y = 0 5)(D 3 + 7D D + 13)y = 0 y(0)=0; y’(0)=2; y’’(0)=-12 6)(4D 4 + 4D 3 – 3D 2 – 2D + 1)y = 0 7)(D 4 – 5D 2 – 6D – 2)y = 0

37 Exercises Find the solution required: 8)(D 3 + D 2 – D – 1)y = 0 y(0)=1; y(2)=0; 9)Find for x = 2 the y value for the particular solution required: (D 3 + 2D 2 )y = 0 y(0)=-3; y’(0)=0; y’’(0)=12


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