First-Order Differential Equations

1
First-Order Differential Equations

• CHAPTER 2

2
Contents

• 2.1 Solution Curves Without a Solution
• 2.2 Separable Variables
• 2.3 Linear Equations
• 2.4 Exact Equations
• 2.5 Solutions by Substitutions
• 2.6 A Numerical Methods
• 2.7 Linear Models
• 2.8 Nonlinear Model
• 2.9 Modeling with Systems of First-Order DEs

3
2.1 Solution Curve Without a Solution

• Introduction Begin our study of first-order DE
  with analyzing a DE qualitatively.
• SlopeA derivative dy/dx of y y(x) gives slopes
  of tangent lines at points.
• Lineal ElementAssume dy/dx f(x, y(x)). The
  value f(x, y) represents the slope of a line, or
  a line element is called a lineal element. See
  Fig2.1

4
Fig2.1

5
Direction Field

• If we evaluate f over a rectangular grid of
  points, and draw a lineal element at each point
  (x, y) of the grid with slope f(x, y), then the
  collection is called a direction field or a slope
  field of the following DE dy/dx f(x, y)

6
Example 1

• The direction field of dy/dx 0.2xy is shown in
  Fig2.2(a) and for comparison with Fig2.2(a), some
  representative graphs of this family are shown in
  Fig2.2(b).

7
Fig2.2
8
Example 2

• Use a direction field to draw an approximate
  solution curve for dy/dx sin y, y(0) -3/2.
• SolutionRecall from the continuity of f(x, y)
  and ?f/?y cos y. Theorem 2.1 guarantees the
  existence of a unique solution curve passing any
  specified points in the plane. Now split the
  region containing (-3/2, 0) into grids. We
  calculate the lineal element of each grid to
  obtain Fig2.3.

9
Fig2.3

10

• Increasing/DecreasingIf dy/dx gt 0 for all x in
  I, then y(x) is increasing in I.If dy/dx lt 0
  for all x in I, then y(x) is decreasing in I.
• DEs Free of the Independent variable
  dy/dx f(y) (1)is
  called autonomous. We shall assume f and f ? are
  continuous on some I.

11
Critical Points

• The zeros of f in (1) are important. If f(c) 0,
  then c is a critical point, equilibrium point or
  stationary point.Substitute y(x) c into (1),
  then we have 0 f(c) 0.
• If c is a critical point, then y(x) c, is a
  solution of (1).
• A constant solution y(x) c of (1) is called an
  equilibrium solution.

12
Example 3

• The following DE dP/dt P(a bP)
• where a and b are positive constants, is
  autonomous.From f(P) P(a bP) 0, the
  equilibrium solutions are P(t) 0 and P(t)
  a/b.
• Put the critical points on a vertical line. The
  arrows in Fig 2.4 indicate the algebraic sign of
  f(P) P(a bP). If the sign is positive or
  negative, then P is increasing or decreasing on
  that interval.

13
Fig2.4
14
Solution Curves

• If we guarantee the existence and uniqueness of
  (1), through any point (x0, y0) in R, there is
  only one solution curve. See Fig 2.5(a).
• Suppose (1) possesses exactly two critical
  points, c1, and c2, where c1 lt c2. The graph of
  the equilibrium solution y(x) c1, y(x) c2 are
  horizontal lines and split R into three regions,
  say R1, R2 and R3 as in Fig 2.5(b).

15
Fig 2.5

16

• Some discussions without proof
• (1) If (x0, y0) in Ri, i 1, 2, 3, when y(x)
  passes through (x0, y0), will remain in the same
  subregion. See Fig 2.5(b).
• (2) By continuity of f , f(y) can not change
  signs in a subregion.
• (3) Since dy/dx f(y(x)) is either positive or
  negative in Ri, a solution y(x) is monotonic.

17

• (4)If y(x) is bounded above by c1, (y(x) lt c1),
  the graph of y(x) will approach y(x) c1If c1
  lt y(x) lt c2, it will approach y(x) c1 and y(x)
  c2If c2 lt y(x) , it will approach y(x)
  c2

18
Example 4

• Referring to example 3, P 0 and P a/b are two
  critical points, so we have three intervals for
  P R1 (-?, 0), R2 (0, a/b), R3 (a/b, ?)
  Let P(0) P0 and when a solution pass through
  P0, we have three kind of graph according to the
  interval where P0 lies on. See Fig 2.6.

19
Fig 2.6
20
Example 5

• The DE dy/dx (y 1)2 possesses the single
  critical point 1. From Fig 2.7(a), we conclude a
  solution y(x) is increasing in -? lt y lt 1 and 1 lt
  y lt ?, where -? lt x lt ?. See Fig 2.7.

21
Fig2.7

22
Attractors and Repellers

• See Fig 2.8(a). When y0 lies on either side of c,
  it will approach c. This kind of critical point
  is said to be asymptotically stable, also called
  an attractor.
• See Fig 2.8(b). When y0 lies on either side of c,
  it will move away from c. This kind of critical
  point is said to be unstable, also called a
  repeller.
• See Fig 2.8(c) and (d). When y0 lies on one side
  of c, it will be attracted to c and repelled from
  the other side. This kind of critical point is
  said to be semi-stable.

23
Fig 2.8

24
Autonomous DEs and Direction Field

• Fig 2.9 shows the direction field of dy/dx 2y
  2.It can be seen that lineal elements passing
  through points on any horizontal line must have
  the same slope. Since the DE has the form dy/dx
  f(y), the slope depends only on y.

25
Fig 2.9
26
2.2 Separable Variables

• Introduction Consider dy/dx f(x, y) g(x).
  The DE dy/dx g(x) (1)can be solved by
  integration. Integrating both sides to get y ?
  g(x) dx G(x) c.eg dy/dx 1 e2x, then y
  ? (1 e2x) dx x ½ e2x c

27

• Rewrite the above equation as
  (2)where p(y) 1/h(y). When h(y) 1,
  (2) reduces to (1).

28

• If y ?(x) is a solution of (2), we must have
  and (3)But dy ? ?(x) dx, (3) is the
  same as (4)

29
Example 1

• Solve (1 x) dy y dx 0.
• SolutionSince dy/y dx/(1 x), we have
  Replacing by c, gives y c(1 x).

30
Example 2

• Solve
• Solution We also can rewrite the solution
  as x2 y2 c2, where c2 2c1Apply the
  initial condition, 16 9 25 c2See Fig2.18.
  

31
Fig2.18
32
Losing a Solution

• When r is a zero of h(y), then y r is also a
  solution of dy/dx g(x)h(y). However, this
  solution will not show up after integration. That
  is a singular solution.

33
Example 3

• Solve dy/dx y2 4.
• SolutionRewrite this DE as (5)then

34
Example 3 (2)

• Replacing exp(c2) by c and solving for y, we
  have (6)If we rewrite the DE as dy/dx
  (y 2)(y 2), from the previous discussion,
  we have y ? 2 is a singular solution.

35
Example 4

• Solve
• SolutionRewrite this DE as using sin 2x 2
  sin x cos x, then ? (ey ye-y) dy 2 ? sin x
  dx from integration by parts, ey ye-y e-y
  -2 cos x c (7)From y(0) 0, we have c 4
  to get ey ye-y e-y 4 -2 cos x (8)

36
Use of Computers

• Let G(x, y) ey ye-y e-y 2 cos x. Using
  some computer software, we plot the level curves
  of G(x, y) c. The resulting graphs are shown in
  Fig2.19 and Fig2.20.

37
Fig2.19 Fig2.20
38

• If we solve dy/dx xy½ , y(0) 0 (9)The
  resulting graphs are shown in Fig2.21.

39
Fig2.21
40
2.3 Linear Equations

• Introduction Linear DEs are friendly to be
  solved. We can find some smooth methods to deal
  with.

41

• Standard FormThe standard for of a first-order
  DE can be written as dy/dx P(x)y f(x) (2)
• The PropertyDE (2) has the property that its
  solution is sum of two solutions, y yc yp,
  where yc is a solution of the homogeneous
  equation dy/dx P(x)y 0 (3)and yp is a
  particular solution of (2).

42

• Verification Now (3) is also separable.
  Rewrite (3) as
• Solving for y gives

43
Variation of Parameters

• Let yp u(x) y1(x), where y1(x) is defined as
  above.We want to find u(x) so that yp is also a
  solution. Substituting yp into (2) gives

44

• Since dy1/dx P(x)y1 0, so that y1(du/dx)
  f(x) Rearrange the above equation, From the
  definition of y1(x), we have (4)

45
Solving Procedures

• If (4) is multiplied by (5)then (
  6)is differentiated (7)we
  get (8)Dividing (8) by
  gives (2).

46
Integrating Factor

• We call y1(x) is an integrating
  factor and we should only memorize this to solve
  problems.

47
Example 1

• Solve dy/dx 3y 6.
• SolutionSince P(x) 3, we have the
  integrating factor is then is the same as
  So e-3xy -2e-3x c, a solution is y -2
  ce-3x, -? lt x lt ?.

48
Notes

• The DE of example 1 can be written as so
  that y 2 is a critical point.

49
General Solutions

• Equation (4) is called the general solution on
  some interval I. Suppose again P and f are
  continuous on I. Writing (2) as Suppose again P
  and f are continuous on I. Writing (2) as y?
  F(x, y) we identify F(x, y) P(x)y f(x),
  ?F/?y P(x)which are continuous on I.Then we
  can conclude that there exists one and only one
  solution of (9)

50
Example 2

• Solve
• SolutionDividing both sides by x, we
  have (10)So, P(x) 4/x, f(x) x5ex,
  P and f are continuous on (0, ?).Since x gt 0, we
  write the integrating factor as

51
Example 2

• Multiply (10) by x-4, Using integration by
  parts, it follows that the general solution on
  (0, ?) is x-4y xex ex c or y x5ex
  x4ex cx4

52
Example 3

• Find the general solution of
• SolutionRewrite as (11)So, P(x)
  x/(x2 9). Though P(x) is continuous on (-?,
  -3), (-3, 3) and (3, ?), we shall solve this DE
  on the first and third intervals. The integrating
  factor is

53
Example 3 (2)

• then multiply (11) by this factor to get and
  Thus, either for x gt 3 or x lt -3, the general
  solution is
• Notes x 3 and x -3 are singular points of
  the DE and is discontinuous at these
  points.

54
Example 4

• Solve
• SolutionWe first have P(x) 1 and f(x) x,
  and are continuous on (-?, ?). The integrating
  factor is , so gives exy
  xex ex c and y x 1 ce-xSince y(0) 4,
  then c 5. The solution is y x 1 5e-x, ?
  lt x lt ? (12)

55

• Notes From the above example, we find yc
  ce-x and yp x 1 we call yc is a transient
  term, since yc ? 0 as x? ?.Some solutions are
  shown in Fig2.24.

Fig2.24
56
Example 5

• Solve , where
• SolutionFirst we see the graph of f(x) in
  Fig2.25.

Fig2.25
57
Example 5 (2)

• We solve this problem on 0 ? x ? 1 and 1 lt x lt
  ?.For 0 ? x ? 1, then y 1 c1e-xSince
  y(0) 0, c1 -1, y 1 - e-xFor x gt 1,
  dy/dx y 0 then y c2e-x

58
Example 5

• We haveFurthermore, we want y(x) is
  continuous at x 1, that is, when x ? 1, y(x)
  y(1) implies c2 e 1.As in Fig2.26, the
  function (13)are continuous on 0,
  ?).

59
Fig2.26
60
Functions Defined by Integrals

• We are interested in the error function and
  complementary error function
  and (14)Since
  , we find erf(x) erfc(x) 1

61
Example 6

• Solve dy/dx 2xy 2, y(0) 1.
• SolutionWe find the integrating factor is
  exp-x2, we get (15)Applying y(0) 1,
  we have c 1. See Fig2.27

62
Fig2.27
63
2.4 Exact Equations

• Introduction Though ydx xdy 0 is separable,
  we can solve it in an alternative way to get the
  implicit solution xy c.

64
Differential of a Function of Two Variables

• If z f(x, y), its differential or total
  differential is (1)Now if z f(x, y)
  c, (2)eg if x2 5xy y3 c, then
  (2) gives (2x 5y) dx (-5x 3y2) dy
  0 (3)

65
(No Transcript)
66
(No Transcript)
67
Proof of Necessity for Theorem 2.1

• If M(x, y) dx N(x, y) dy is exact, there
  exists some function f such that for all x in R
  M(x, y) dx N(x, y) dy (?f/?x) dx (?f/?y)
  dyTherefore M(x, y) , N(x, y)
  and The sufficient part consists of showing
  that there is a function f for which
  M(x, y) and N(x, y)

68
Method of Solution

• Since ?f/?x M(x, y), we have
  (5)Differentiating (5) with respect to y and
  assume ?f/?y N(x, y)Then and
  (6)

69

• Integrate (6) with respect to y to get g(y), and
  substitute the result into (5) to obtain the
  implicit solution f(x, y) c.

70
Example 1

• Solve 2xy dx (x2 1) dy 0.
• SolutionWith M(x, y) 2xy, N(x, y) x2 1,
  we have ?M/?y 2x ?N/?xThus it is exact.
  There exists a function f such that ?f/?x
  2xy, ?f/?y x2 1Then f(x, y) x2y
  g(y) ?f/?y x2 g(y) x2 1 g(y) -1,
  g(y) -y

71
Example 1 (2)

• Hence f(x, y) x2y y, and the solution
  is x2y y c, y c/(1 x2)The interval of
  definition is any interval not containing x 1
  or x -1.

72
Example 2

• Solve (e2y y cos xy)dx(2xe2y x cos xy
  2y)dy 0.
• SolutionThis DE is exact because ?M/?y 2e2y
  xy sin xy cos xy ?N/?xHence a function f
  exists, and ?f/?y 2xe2y x cos xy 2ythat
  is,

73
Example 2 (2)

• Thus h(x) 0, h(x) c. The solution is xe2y
  sin xy y2 c 0

74
Example 3

• Solve
• SolutionRewrite the DE in the form (cos x sin
  x xy2) dx y(1 x2) dy 0 Since ?M/?y
  2xy ?N/?x (This DE is exact)Now ?f/?y y(1
  x2) f(x, y) ½y2(1 x2) h(x) ?f/?x
  xy2 h(x) cos x sin x xy2

75
Example 3 (2)

• We have h?(x) cos x sin x h(x) -½ cos2
  xThus ½y2(1 x2) ½ cos2 x c1 or y2(1
  x2) cos2 x c (7)where c 2c1. Now
  y(0) 2, so c 3.The solution is y2(1 x2)
  cos2 x 3

76
Fig 2.28

• Fig 2.28 shows the family curves of the above
  example and the curve of the specialized VIP is
  drawn in color.

77
Integrating Factors

• It is sometimes possible to find an integrating
  factor?(x, y), such that ?(x, y)M(x, y)dx
  ?(x, y)N(x, y)dy 0 (8)is an exact
  differential.Equation (8) is exact if and only
  if (?M)y (?N)x Then ?My ?yM ?Nx
  ?xN, or ?xN ?yM (My Nx) ? (9)

78

• Suppose ? is a function of one variable, say x,
  then ?x d? /dx(9) becomes (10)If
  we have (My Nx) / N depends only on x, then
  (10) is a first-order ODE and is separable.
  Similarly, if ? is a function of y only, then
  (11)In this case, if (Nx My) / M is
  a function of y only, then we can solve (11) for
  ?.

79

• We summarize the results for M(x, y) dx N(x,
  y) dy 0 (12)If (My Nx) / N depends only on
  x, then (13)If (Nx My) / M depends
  only on y, then (14)

80
Example 4

• The nonlinear DE xy dx (2x2 3y2 20) dy 0
  is not exact. With M xy, N 2x2 3y2 20, we
  find My x, Nx 4x. Since depends on both x
  and y. depends only on y.The integrating
  factor is e ? 3dy/y e3lny y3 ?(y)

81
Example 4 (2)

• then the resulting equation is xy4 dx (2x2y3
  3y5 20y3) dy 0It is left to you to verify
  the solution is ½ x2y4 ½ y6 5y4 c

82
2.5 Solutions by Substitutions

• IntroductionIf we want to transform the
  first-order DE dx/dy f(x, y)by the
  substitution y g(x, u), where u is a function
  of x, then Since dy/dx f(x, y), y g(x,
  u), Solving for du/dx, we have the form du/dx
  F(x, u). If we can get u ?(x), a solution is
  y g(x, ?(x)).

83
Homogeneous Equations

• If a function f has the property f(tx, ty)
  t?f(x, y), then f is called a homogeneous
  function of degree ?.eg f(x, y) x3 y3 is
  homogeneous of degree 3, f(tx, ty) (tx)3
  (ty)3 t3f(x, y)
• A first-order DE M(x, y) dx N(x, y) dy
  0 (1)is said to be homogeneous, if both M and
  N are homogeneous of the same degree, that is,
  if M(tx, ty) t?M(x, y), N(tx, ty) t?N(x, y)

84

• Note Here the word homogeneous is not the same
  as in Sec 2.3.
• If M and N are homogeneous of degree ?, M(x,
  y)x? M(1, u), N(x, y)x?N(1, u), uy/x (2)M(x,
  y)y? M(v, 1), N(x, y)y?N(v, 1), vx/y (3)Then
  (1) becomes x? M(1, u) dx x? N(1, u) dy 0,
  or M(1, u) dx N(1, u) dy 0where u y/x
  or y ux and dy udx xdu,

85

• then M(1, u) dx N(1, u)(u dx x du) 0,
  and M(1, u) u N(1, u) dx xN(1, u) du
  0or

86
Example 1

• Solve (x2 y2) dx (x2 xy) dy 0.
• Solution We have M x2 y2, N x2 xy are
  homogeneous of degree 2. Let y ux, dy u dx
  x du, then (x2 u2x2) dx (x2 - ux2)(u dx x
  du) 0

87
Example 1 (2)

• Then Simplify to get
• Note We may also try x vy.

88
Bernoullis Equation

• The DE dy/dx P(x)y f(x)yn (4)where n is
  any real number, is called Bernoullis Equation.
• Note for n 0 and n 1, (4) is linear,
  otherwise, let u y1-n to reduce (4) to a
  linear equation.

89
Example 2

• Solve x dy/dx y x2y2.
• SolutionRewrite the DE as dy/dx (1/x)y
  xy2With n 2, then y u-1, and dy/dx
  -u-2(du/dx) From the substitution and
  simplification, du/dx (1/x)u -xThe
  integrating factor on (0, ?) is

90
Example 2 (2)

• Integrating gives x-1u -x c, or u -x2
  cx. Since u y-1, we have y 1/u and a
  solution of the DE is y 1/(-x2 cx).

91
Reduction to Separation of Variables

• A DE of the form dy/dx f(Ax By
  C) (5)can always be reduced to a separable
  equation by means of substitution u Ax By
  C.

92
Example 3

• Solve dy/dx (-2x y)2 7, y(0) 0.
• SolutionLet u -2x y, then du/dx -2
  dy/dx, du/dx 2 u2 7 or du/dx u2
  9This is separable. Using partial fractions,
  or

93
Example 3 (2)

• then we have or Solving the equation for u
  and the solution is or (6)Applying
  y(0) 0 gives c -1.

94
Example 3 (3)

• The graph of the particular solution is shown
  in Fig 2.30 in solid color.

95
Fig 2.30

96
2.6 A Numerical Method

• Using the Tangent LineLet us assume y f(x,
  y), y(x0) y0 (1)possess a solution. For
  example,
  the resulting graph is shown in Fig 2.31.
  

97
Fig 2.31
98
Eulers Method

• Using the linearization of the unknown solution
  y(x) of (1) at x0, L(x) f(x0, y0)(x - x0)
  y0 (2)Replacing x by x1 x0 h, we
  have L(x1) f(x0, y0)(x0 h - x0) y0 or
  y1 y0 h f(x0, y0)and yn1 yn h f(xn,
  yn) (3)where xn x0 nh. See Fig 2.32

99
Fig 2.32
100
Example 1

• Consider Use Eulers method to obtain
  y(2.5) using h 0.1 and then h 0.05.
• SolutionLet
  the results step by step are shown in Table
  2.1 and table 2.2.

101
Table 2.1 Table 2.2
Table 2.1 h 0.1 Table 2.1 h 0.1
xn yn
2.00 4.0000
2.10 4.1800
2.20 4.3768
2.30 4.5914
2.40 4.8244
2.50 5.0768
Table 2.2 h 0.05 Table 2.2 h 0.05
xn yn
2.00 4.0000
2.05 4.0900
2.10 4.1842
2.15 4.2826
2.20 4.3854
2.25 4.4927
2.30 4.6045
2.35 4.7210
2.40 4.8423
2.45 4.9686
2.50 5.0997
102
Example 2

• Consider y 0.2xy, y(1) 1. Use Eulers method
  to obtain y(1.5) using h 0.1 and then h 0.05.
• SolutionWe have f(x, y) 0.2xy, the results
  step by step are shown in Table 2.3 and table
  2.4.

103
Table 2.3
Table 2.3 h 0.1 Table 2.3 h 0.1 Table 2.3 h 0.1 Table 2.3 h 0.1 Table 2.3 h 0.1
xn yn ActualValue AbsoluteError Rel.Error
1.00 1.0000 1.0000 0.0000 0.00
1.10 1.0200 1.0212 0.0012 0.12
1.20 1.0424 1.0450 0.0025 0.24
1.30 1.0675 1.0714 0.0040 0.37
1.40 1.0952 1.1008 0.0055 0.50
1.50 1.1259 1.1331 0.0073 0.64
104
Table 2.4
Table 2.4 h 0.05 Table 2.4 h 0.05 Table 2.4 h 0.05 Table 2.4 h 0.05 Table 2.4 h 0.05
xn yn ActualValue AbsoluteError Rel.Error
1.00 1.0000 1.0000 0.0000 0.00
1.05 1.0100 1.0103 0.0003 0.03
1.10 1.0206 1.0212 0.0006 0.06
1.15 1.0318 1.0328 0.0009 0.09
1.20 1.0437 1.0450 0.0013 0.12
1.25 1.0562 1.0579 0.0016 0.16
1.30 1.0694 1.0714 0.0020 0.19
1.35 1.0833 1.0857 0.0024 0.22
1.40 1.0980 1.1008 0.0028 0.25
1.45 1.1133 1.1166 0.0032 0.29
1.50 1.1295 1.1331 0.0037 0.32
105
Numerical Solver

• See Fig 2.33 to know the comparisons of numerical
  methods.

Fig 2.33
106
Using a Numerical Solver

• When the result is not helpful by numerical
  solvers, as in Fig 2.34, we may decrease the step
  size, use another method, or use another
  solver.

Fig 2.34
107
2.7 Linear Models

• Growth and Decay (1)

108
Example 1 Bacterial Growth

• P0 initial number of bacterial P(0)P(1)
  3/2 P(0)Find the time necessary for triple
  number.
• SolutionSince dP/dt kt, dP/dt kt 0, we
  have P(t) cekt, using P(0) P0then c P0
  and P(t) P0ekt Since P(1) 3/2 P(0), then
  P(1) P0ek 3/2 P(0)So, k ln(3/2)
  0.4055.Now P(t) P0e0.4055t 3P0 , t
  ln3/0.4055 2.71.See Fig 2.35.

109
Fig 2.35

110
Fig 2.36

• k gt 0 is called a growth constant, and k gt 0 is
  called a decay constant. See Fig 2.36.

111
Example 2 Half-Life of Plutonium

• A reactor converts U-238 into the isotope
  plutonium-239. After 15 years, there is 0.043 of
  the initial amount A0 of the plutonium has
  disintegrated. Find the half-life of this
  isotope.
• SolutionLet A(t) denote the amount of plutonium
  remaining at time t. The DE is as (2)The
  solution is A(t) A0ekt. If 0.043 of A0 has
  disintegrated, then 99.957 remains.

112
Example 2 (2)

• Then, 0.99957A0 A(15) A0e15k, then k (ln
  0.99957) / 15 -0.00002867 Let A(t)
  A0e-0.00002867t ½ A0Then

113
Example 3 Carbon Dating

• A fossilized bone contains 1/1000 the original
  amount C-14. Determine the age of the fossil.
• SolutionWe know the half-life of C-14 is 5600
  years.Then A0 /2 A0e5600k, k -(ln 2)/5600
  -0.00012378.And A(t) A0 /1000 A0e
  -0.00012378t

114
Newtons Law of Cooling

• (3)where Tm is the temperature of the
  medium around the object.

115
Example 4

• A cakes temperature is 300?F. Three minutes
  later its temperature is 200?F. How long will it
  for this cake to cool off to a room temperature
  of 70?F?
• SolutionWe identify Tm 70, then (4)a
  nd T(3) 200. From (4), we have

116
Example 4 (2)

• Using T(0) 300 then c2 230Using T(3) 200
  then e3k 13/23, k -0.19018Thus T(t) 70
  230e-0.19018t (5)From (5), only t ?, T(t)
  70. It means we need a reasonably long time to
  get T 70. See Fig 2.37.

117
Fig 2.37
118
Mixtures

• (6)

119
Example 5

• Recall from example 5 of Sec 1.3, we have
  How much salt is in the tank after along
  time?
• SolutionSince Using x(0) 50, we have
  x(t) 600 - 550e-t/100 (7)When t is large
  enough, x(t) 600.

120
Fig 2.38

121
Series Circuits

• See Fig 2.39. (8)See Fig
  2.40. (9) (10)

122
Fig 2.39
123
Fig2.40
124
Example 6

• Refer to Fig 2.39, where E(t) 12 Volt, L ½
  HenryR 10 Ohms. Determine i(t) where i(0) 0.
• SolutionFrom (8), Then Using i(0) 0,
  c -6/5, then i(t) (6/5) (6/5)e-20t.

125
Example 6 (2)

• A general solution of (8) is (11)When
  E(t) E0 is a constant, (11) becomes
  (12)where the first term is called a
  steady-state part, and the second term is a
  transient term.

126
Note

• Referring to example 1, P(t) is a continuous
  function. However, it should be discrete. Keeping
  in mind, a mathematical model is not reality.
  See Fig 2.41.

127
Fig 2.41

128
2.8 Nonlinear Models

• Population DynamicsIf P(t) denotes the size of
  population at t, the relative (or specific),
  growth rate is defined by (1)When a
  population growth rate depends on the present
  number , the DE is (2)which is called
  density-dependent hypothesis.

129
Logistic Equation

• If K is the carrying capacity, from (2) we have
  f(K) 0, and simply set f(0) r. Fig 2.46
  shows three functions that satisfy these two
  conditions.

130
Fig 2.46
131

• Suppose f (P) c1P c2. Using the conditions,
  we have c2 r, c1 -r/K. Then (2) becomes
  (3)Relabel (3), then (4)whi
  ch is known as a logistic equation, its solution
  is called the logistic function and its graph is
  called a logistic curve.

132
Solution of the Logistic Equation

• From After simplification, we have

133

• If P(0) P0 ? a/b, then c1 P0/(a
  bP0) (5)

134
Graph of P(t)

• Form (5), we have the graph as in Fig 2.47. When
  0 lt P0 lt a/2b, see Fig 2.47(a).When a/2b lt P0 lt
  a/b, see Fig 2.47(b).

Fig 2.47
135
Example 1

• Form the previous discussion, assume an isolated
  campus of 1000 students, then we have the
  DE Determine x(6).
• SolutionIdentify a 1000k, b k, from (5)

136
Example 1 (2)

• Since x(4) 50, then -1000k -0.9906,
  Thus x(t) 1000/(1 999e-0.9906t)
• See Fig 2.48.

137
Fig 2.48

138
Modification of the Logistic Equation

• or (6)or (7)which is
  known as the Gompertz DE.

139
Chemical Reactions

• (8)or (9)

140
Example 2

• The chemical reaction is described as
  Then By separation of variables and partial
  fractions, (10)Using X(10) 30, 210k
  0.1258, finally (11)See Fig 2.49.

141
Fig 2.49

142
2.9 Modeling with Systems of First-Order DEs

• Systems (1)where g1 and g2 are linear
  in x and y.
• Radioactive Decay Series (2)

143
Mixtures

• From Fig 2.52, we have (3)

144
Fig 2.52
145
A Predator-Prey Model

• Let x, y denote the fox and rabbit populations at
  t.When lacking of food, dx/dt ax, a gt
  0 (4)When rabbits are present, dx/dt
  ax bxy (5)When lacking of foxes, dy/dt
  dy, d gt 0 (6) When foxes are present,
  dy/dt dy cxy (7)

146

• Then (8)which is known as the
  Lotka-Volterra predator-prey model.

147
Example 1

• Suppose Figure 2.53 shows the graph of the
  solution.

148
Fig 2.53

149
Competition Models

• dx/dt ax, dy/dt cy (9)Two species
  compete, then dx/dt ax by dy/dt cy
  dx (10)or dx/dt ax bxy dy/dt cy
  dxy (11)or dx/dt a1x b1x2 dy/dt
  a2y b2y2 (12)

150

• or dx/dt a1x b1x2 c1xy dy/dt a2y
  b2y2 c2xy (13)

151
Network

• Referring to Fig 2.54, we have i1(t) i2(t)
  i3(t) (14) (15) (16)

152

• Using (14) to eliminate i1, then (17)
  Referring to Fig 2.55, please verify
  (18)

153
Fig 2.54

154
Fig 2.55
155
Thank You !