1. Section 5.1 First-Order Systems & Applications

2. Suppose x and y are both functions of t. Solve:
x′ = 3x – y
y′ = 2x + y – et

3. Why would we consider such things?

4. Ex. 1 Give the system of diff eqs which describe the following tanks containing brine solution:

5. Theorem:
Consider the following system of diff eqs (all xi, pij, and fi are functions of t )
x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
x2′ = p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2
x3′ = p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3
: :
xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn
Let J be an open interval containing t = a. Suppose the functions pij and the functions fk are continuous on J. Then the system of differential equations has a unique solution that satisfies the following initial conditions:
x1(a) = b1 , x2(a) = b2 , x3(a) = b3 , , xn(a) = bn

6. If we have a system of higher order diff eqs, we can transform it into a system of first order diff eqs.

7. Ex. 3 Rewrite the one diff eq x(3) + 3x″ + 2x′ – 5x = sin(2t) as a system of first order diff eqs.

8. Ex. 4 Rewrite the following system as a system of first order diff eqs.
2x″ = –6x + 2y
y″ = 2x – 2y + 40sin(3t)

9. Section 5.2 The Method Of Elimination

10. Review of solving systems of algebraic equations (2 equations with 2 unknowns)
i. Substitution method
ii. Elimination method
iii. Cramer's rule

11. Ex. 1 Find the general solution to the following system by using a variant of the substitution method.
x′ = 4x – 3y
y′ = 6x – 7y

12. Ex. 1 Find the general solution to the following system by using a variant of the substitution method.
x′ = 4x – 3y
y′ = 6x – 7y

13. Ex. 2 Find the general solution to the following system by using a variant of the elimination method.
x′ = 4x – 3y
y′ = 6x – 7y

14. Ex. 2 Find the general solution to the following system by using a variant of the elimination method.
x′ = 4x – 3y
y′ = 6x – 7y

15. We shall now use the following notation:
L will be a linear operator of the form
L = anDn + an–1Dn–1 + an–2Dn–2 + ⋯ + a2D2 + a1D + a0

16. Ex. 3 Find the general solution to the following general system of diff eqs:
L1x + L2y = f1(t)
L3x + L4y = f2(t)

17. Ex. 3 Find the general solution to the following general system of diff eqs:
L1x + L2y = f1(t)
L3x + L4y = f2(t)

18. Ex. 4 Find the general solution to x′ = 2x + y
y′ = 2x + 3y + e5t
Solution: (D–2)x – y = 0
–2x + (D–3)y = e5t
[(D–2)(D–3) – 2] x = (D – 3)0 + e5t [(D–2)(D–3) – 2] y = (D – 2) e5t + 2(0)
(D2–5D+4) x = e5t (D2–5D+4) y = 5e5t – 2e5t
(D–4)(D–1) x = e5t (D–4)(D–1) y = 3e5t
xc = c1e4t + c2et yc = c3e4t + c4et
xp = Ae5t ⇒ xp = (1∕4)e5t yp = Be5t ⇒ yp = (3∕4)e5t
x = c1e4t + c2et + (1∕4)e5t y = c3e4t + c4et + (3∕4)e5t 

19. Ex. 4 Find the general solution to x′ = 2x + y
y′ = 2x + 3y + e5t
Solution:
x = c1e4t + c2et + (1∕4)e5t y = c3e4t + c4et + (3∕4)e5t
Plugging these solutions into the first equation in the initial problem we get:
x′ = 2x + y
(c1e4t + c2et + (1∕4)e5t)′ = 2(c1e4t + c2et + (1∕4)e5t) + (c3e4t + c4et + (3∕4)e5t)
4c1e4t + c2et + (5/4)e5t = 2c1e4t + 2c2et + (1/2)e5t + c3e4t + c4et + (3/4)e5t
0 = (–2c1 + c3)e4t + (c2 + c4)et
–2c1 + c3 = 0 ⇒ c3 = 2c1
c2 + c4 = 0 c4 = –c2

20. Section 5.3 Matrices & Linear Systems

21. x1′ = 2x1 + x2
x2′ = 2x1 + 3x2

22. x1′ = 2x1 + x2 ⇒
x2′ = 2x1 + 3x2

23. Previously we would state:
The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et
x2′ = 2x1 + 3x x2 = 2c1e4t – c2et  

24. Previously we would state:
The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et
x2′ = 2x1 + 3x x2 = 2c1e4t – c2et  
We now would state:
The general solution to the system turns out to be:

25. Previously we would state:
The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et
x2′ = 2x1 + 3x x2 = 2c1e4t – c2et  
We now would state:
The general solution to the system turns out to be:
We shall find that general solutions to these "2x2 homogenous systems" will take this form of (where and are two linearly independent vectors).

26. x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
: :
xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn

27. x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
: :
xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn

28. x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
: :
xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn
This system of diff eqs is said to be homogenous if

29. x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1
: :
xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn
This system of diff eqs is said to be homogenous if
Thus, the matrix equation would be for a homogenous system.

30. We shall now see many theorems very similar to theorems and definitions we had in chapter 2.

31. are said to be linearly independent if the equation
Definition:
are said to be linearly independent if the equation
only has the solution of c1 = c2 = ⋯ = cn = 0.

32. Definition:
If are solutions to then we define the Wronskian of these solution to be

33. Theorem:
Suppose are solutions to on an interval J where all the pij functions are continuous.
(a) If are linearly dependent then W = 0 for every point on the interval J.
(b) If are linearly independent then W ≠ 0 for every point on the interval J.

34. Theorem:
Suppose are linearly independent solutions to on an interval J where all the pij functions are continuous. The general solution to is given as:

35. Theorem:
Suppose is a particular solution to the nonhomogenous system
and is the general solution to the corresponding homogenous system Then the general solution to the nonhomogenous system is

36. Ex. 1 x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 = x2 – x3
(a) Write this system as a matrix equation.

37. Ex. 1 x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 = x2 – x3
(b) Verify that the following three
vectors are solution vectors:

38. Ex. 1 x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 = x2 – x3
(c) Verify that these are
linearly independent vectors.

39. Ex. 1 x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 = x2 – x3
(d) Give the general solution for the system.

40. Ex. 1 x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 = x2 – x3
(e) Give the general solution for x1.
Give the general solution for x2.
Give the general solution for x3.

41. Ex. 1 x′1 = x1 + x2 – 2x3
x′2 = –x1 + 2x2 + x3
x′3 = x2 – x3
(f) Solve the initial value problem:
x′1 = x1 + x2 – 2x3 x1(0) = 9, x2(0) = –2, x3(0) = 5
x′3 = x2 – x3

42. Section 5.4 The Eigenvalue Method for Homogeneous Systems

43. Review of eigenvalues and eigenvectors:
Let A be a square matrix. The vector is said to be an eigenvector for the eigenvalue λ if

44. Review of eigenvalues and eigenvectors:
Let A be a square matrix. The vector is said to be an eigenvector for the eigenvalue λ if
What is the connection between eigenvectors and solutions to systems of diff eqs?

45. Ex. 1 We previously (in section 5
Ex. 1 We previously (in section 5.3) found three linearly independent solution
vectors to the system One of these was
Verify that this solution is an eigenvector of

46. Ex. 2 Suppose that (where A, B, C, and k are constants) is a solution
vector to the system Show that this solution vector is an eigenvector of the matrix P.

47. Theorem:
Given a homogenous system , suppose λ is an eigenvalue of P with eigenvector Then is a nontrivial solution vector to

48. Ex. 3 Use eigenvalues/eigenvectors to find the general solution to the following system and write your final solution in scalar form.
x′1 = 3x1 + x2
x′2 = 3x1 + 5x2

49. Ex. 3 Use eigenvalues/eigenvectors to find the general solution to the following system and write your final solution in scalar form.
x′1 = 3x1 + x2
x′2 = 3x1 + 5x2

50. Note:
Suppose where P is an nxn matrix. If there are n distinct real eigenvalues of P then we've got n linearly independent solution vectors. This means we can write down the general solution as
(here the λi are the eigenvalues, the are the eigenvectors and the ci are arbitrary constants).
If we have less than n distinct eigenvalues, or if some of the eigenvalues are complex then we will run into trouble and need to do something else.

51. Theorem:
Given a homogenous system , suppose α + βi is a complex eigenvalue of P with eigenvector Then the real and
imaginary parts of the vector will form two linearly independent solution vectors.

52. Theorem:
Given a homogenous system , suppose α + βi is a complex eigenvalue of P with eigenvector Then the real and
imaginary parts of the vector will form two linearly independent solution vectors.
Note that there are formulas for these two vectors, but you should not use them!
Instead of using these formulas we shall just use this theorem which indicates that we should find the one vector , then split it up into its real and imaginary parts.

53. Ex. 4 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –x1 – x2
x′2 = 4x1 – x2

54. Ex. 4 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –x1 – x2
x′2 = 4x1 – x2

55. Section 5.5 Multiple Eigenvalue Solutions

56. If we have less than n distinct eigenvalues (where P is an nxn matrix) then we may have a hard time finding enough linearly independent solution vectors. When an eigenvalue, say λ, is a repeated root with multiplicity k of the characteristic equation, then two possibilities arise:

57. If we have less than n distinct eigenvalues (where P is an nxn matrix) then we may have a hard time finding enough linearly independent solution vectors. When an eigenvalue, say λ, is a repeated root with multiplicity k of the characteristic equation, then two possibilities arise:
1. There are "just enough" linearly independent eigenvectors associated with
. Then there are "just enough" solution vectors:
for us to form the general solution to the diff eqs.
2. There are not enough linearly independent eigenvectors associated with λ.
In this case λ is said to be defective.
We shall cover the first of these two cases.

58. Ex. 1 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –13x1 + 40x2 – 48x3
x′2 = –8x1 + 23x2 – 24x3
x′3 = x3

59. Ex. 1 Use eigenvalues/eigenvectors to find the general solution to
x′1 = –13x1 + 40x2 – 48x3
x′2 = –8x1 + 23x2 – 24x3
x′3 = x3