First-Order Differential Equations CHAPTER 2

CH2_2 Contents  2.1 Solution Curves Without a Solution 2.1 Solution Curves Without a Solution  2.2 Separable Variables 2.2 Separable Variables  2.3 Linear Equations 2.3 Linear Equations  2.4 Exact Equations 2.4 Exact Equations  2.5 Solutions by Substitutions 2.5 Solutions by Substitutions  2.6 A Numerical Methods 2.6 A Numerical Methods  2.7 Linear Models 2.7 Linear Models  2.8 Nonlinear Model 2.8 Nonlinear Model  2.9 Modeling with Systems of First-Order DEs 2.9 Modeling with Systems of First-Order DEs

CH2_3 2.1 Solution Curve Without a Solution  Introduction: Begin our study of first-order DE with analyzing a DE qualitatively.  Slope A derivative dy/dx of y = y(x) gives slopes of tangent lines at points.  Lineal Element Assume 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

CH2_4 Fig2.1

CH2_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)

CH2_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).

CH2_7 Fig2.2

CH2_8 Example 2 Use a direction field to draw an approximate solution curve for dy/dx = sin y, y(0) = −3/2. Solution: Recall 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.

CH2_9 Fig2.3

CH2_10  Increasing/Decreasing If dy/dx > 0 for all x in I, then y(x) is increasing in I. If dy/dx < 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.

CH2_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.

CH2_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.

CH2_13 Fig2.4

CH2_14 Solution Curves  If we guarantee the existence and uniqueness of (1), through any point (x 0, y 0 ) in R, there is only one solution curve. See Fig 2.5(a).  Suppose (1) possesses exactly two critical points, c 1, and c 2, where c 1 < c 2. The graph of the equilibrium solution y(x) = c 1, y(x) = c 2 are horizontal lines and split R into three regions, say R 1, R 2 and R 3 as in Fig 2.5(b).

CH2_15 Fig 2.5

CH2_16  Some discussions without proof: (1) If (x 0, y 0 ) in R i, i = 1, 2, 3, when y(x) passes through (x 0, y 0 ), 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 R i, a solution y(x) is monotonic.

CH2_17 (4)If y(x) is bounded above by c 1, (y(x) < c 1 ), the graph of y(x) will approach y(x) = c 1 ; If c 1 < y(x) < c 2, it will approach y(x) = c 1 and y(x) = c 2 ; If c 2 < y(x), it will approach y(x) = c 2 ;

CH2_18 Example 4 Referring to example 3, P = 0 and P = a/b are two critical points, so we have three intervals for P: R 1 : (- , 0), R 2 : (0, a/b), R 3 : (a/b,  ) Let P(0) = P 0 and when a solution pass through P 0, we have three kind of graph according to the interval where P 0 lies on. See Fig 2.6.

CH2_19 Fig 2.6

CH2_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 -  < y < 1 and 1 < y < , where -  < x < . See Fig 2.7.

CH2_21 Fig2.7

CH2_22 Attractors and Repellers  See Fig 2.8(a). When y 0 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 y 0 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 y 0 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.

CH2_23 Fig 2.8

CH2_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.

CH2_25 Fig 2.9

CH2_ 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 + e 2x, then y =  (1 + e 2x ) dx = x + ½ e 2x + c A first-order DE of the form dy/dx = g(x)h(y) is said to be separable or to have separable variables. DEFINITION 2.1 Separable Equations

CH2_27  Rewrite the above equation as (2) where p(y) = 1/h(y). When h(y) = 1, (2) reduces to (1).

CH2_28  If y =  (x) is a solution of (2), we must have and (3) But dy =  (x) dx, (3) is the same as (4)

CH2_29 Example 1 Solve (1 + x) dy – y dx = 0. Solution: Since dy/y = dx/(1 + x), we have Replacing by c, gives y = c(1 + x).

CH2_30 Example 2 Solve Solution: We also can rewrite the solution as x 2 + y 2 = c 2, where c 2 = 2c 1 Apply the initial condition, = 25 = c 2 See Fig2.18.

CH2_31 Fig2.18

CH2_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.

CH2_33 Example 3 Solve dy/dx = y 2 – 4. Solution: Rewrite this DE as (5) then

CH2_34 Example 3 (2) Replacing exp(c 2 ) 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.

CH2_35 Example 4 Solve Solution: Rewrite this DE as using sin 2x = 2 sin x cos x, then  (e y – ye -y ) dy = 2  sin x dx from integration by parts, e y + ye -y + e -y = -2 cos x + c(7) From y(0) = 0, we have c = 4 to get e y + ye -y + e -y = 4 −2 cos x(8)

CH2_36 Use of Computers  Let G(x, y) = e y + 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.

CH2_37 Fig2.19 Fig2.20

CH2_38  If we solve dy/dx = xy ½, y(0) = 0(9) The resulting graphs are shown in Fig2.21.

CH2_39 Fig2.21

CH2_ Linear Equations  Introduction: Linear DEs are friendly to be solved. We can find some smooth methods to deal with. A first-order DE of the form a 1 (x)(dy/dx) + a 0 (x)y = g(x)(1) is said to be a linear equation in y. When g(x) = 0, (1) is said to be homogeneous; otherwise it is nonhomogeneous. DEFINITION 2.2 Linear Equations

CH2_41  Standard Form The standard for of a first-order DE can be written as dy/dx + P(x)y = f(x)(2)  The Property DE (2) has the property that its solution is sum of two solutions, y = y c + y p, where y c is a solution of the homogeneous equation dy/dx + P(x)y = 0(3) and y p is a particular solution of (2).

CH2_42  Verification Now (3) is also separable. Rewrite (3) as  Solving for y gives

CH2_43  Let y p = u(x) y 1 (x), where y 1 (x) is defined as above. We want to find u(x) so that y p is also a solution. Substituting y p into (2) gives Variation of Parameters

CH2_44  Since dy 1 /dx + P(x)y 1 = 0, so that y 1 (du/dx) = f(x) Rearrange the above equation, From the definition of y 1 (x), we have (4)

CH2_45 Solving Procedures  If (4) is multiplied by (5) then (6) is differentiated (7) we get (8) Dividing (8) by gives (2).

CH2_46  We call y 1 (x) = is an integrating factor and we should only memorize this to solve problems. Integrating Factor

CH2_47 Example 1 Solve dy/dx – 3y = 6. Solution: Since P(x) = – 3, we have the integrating factor is then is the same as So e -3x y = -2e -3x + c, a solution is y = -2 + ce -3x, -  < x < .

CH2_48  The DE of example 1 can be written as so that y = –2 is a critical point. Notes

CH2_49  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) General Solutions

CH2_50 Example 2 Solve Solution: Dividing both sides by x, we have (10) So, P(x) = –4/x, f(x) = x 5 e x, P and f are continuous on (0,  ). Since x > 0, we write the integrating factor as

CH2_51 Example 2 Multiply (10) by x -4, Using integration by parts, it follows that the general solution on (0,  ) is x -4 y = xe x – e x + c or y = x 5 e x – x 4 e x + cx 4

CH2_52 Example 3 Find the general solution of Solution: Rewrite as (11) So, P(x) = x/(x 2 – 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

CH2_53 Example 3 (2) then multiply (11) by this factor to get and Thus, either for x > 3 or x < -3, the general solution is  Notes: x = 3 and x = -3 are singular points of the DE and is discontinuous at these points.

CH2_54 Example 4 Solve Solution: We first have P(x) = 1 and f(x) = x, and are continuous on (- ,  ). The integrating factor is, so gives e x y = xe x – e x + c and y = x – 1 + ce -x Since y(0) = 4, then c = 5. The solution is y = x – 1 + 5e -x, –  < x <  (12)

CH2_55  Notes: From the above example, we find y c = ce -x and y p = x – 1 we call y c is a transient term, since y c  0 as x  . Some solutions are shown in Fig2.24. Fig2.24

CH2_56 Example 5 Solve, where Solution: First we see the graph of f(x) in Fig2.25. Fig2.25

CH2_57 Example 5 (2) We solve this problem on 0  x  1 and 1 1, dy/dx + y = 0 then y = c 2 e -x

CH2_58 Example 5 We have Furthermore, we want y(x) is continuous at x = 1, that is, when x  1 +, y(x) = y(1) implies c 2 = e – 1. As in Fig2.26, the function (13) are continuous on [0,  ).

CH2_59 Fig2.26

CH2_60  We are interested in the error function and complementary error function and(14) Since, we find erf(x) + erfc(x) = 1 Functions Defined by Integrals

CH2_61 Example 6 Solve dy/dx – 2xy = 2, y(0) = 1. Solution: We find the integrating factor is exp{-x 2 }, we get(15) Applying y(0) = 1, we have c = 1. See Fig2.27

CH2_62 Fig2.27

CH2_ Exact Equations  Introduction: Though ydx + xdy = 0 is separable, we can solve it in an alternative way to get the implicit solution xy = c.

CH2_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 x 2 – 5xy + y 3 = c, then (2) gives (2x – 5y) dx + (-5x + 3y 2 ) dy = 0(3)

CH2_65 An expression M(x, y) dx + N(x, y) dy is an exact differential in a region R corresponding to the differential of some function f(x, y). A first-order DE of the form M(x, y) dx + N(x, y) dy = 0 is said to be an exact equation, if the left side is an exact differential. DEFINITION 2.3 Exact Equation

CH2_66 Let M(x, y) and N(x, y) be continuous and have continuous first partial derivatives in a region R defined by a < x < b, c < y < d. Then a necessary and sufficient condition that M(x, y) dx + N(x, y) dy be an exact differential is (4) THEOREM 2.1 Criterion for an Extra Differential

CH2_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) dy Therefore 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)

CH2_68  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) Method of Solution

CH2_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.

CH2_70 Example 1 Solve 2xy dx + (x 2 – 1) dy = 0. Solution: With M(x, y) = 2xy, N(x, y) = x 2 – 1, we have  M/  y = 2x =  N/  x Thus it is exact. There exists a function f such that  f/  x = 2xy,  f/  y = x 2 – 1 Then f(x, y) = x 2 y + g(y)  f/  y = x 2 + g’(y) = x 2 – 1 g’(y) = -1, g(y) = -y

CH2_71 Example 1 (2) Hence f(x, y) = x 2 y – y, and the solution is x 2 y – y = c, y = c/(1 – x 2 ) The interval of definition is any interval not containing x = 1 or x = -1.

CH2_72 Example 2 Solve (e 2y – y cos xy)dx+(2xe 2y – x cos xy + 2y)dy = 0. Solution: This DE is exact because  M/  y = 2e 2y + xy sin xy – cos xy =  N/  x Hence a function f exists, and  f/  y = 2xe 2y – x cos xy + 2y that is,

CH2_73 Example 2 (2) Thus h’(x) = 0, h(x) = c. The solution is xe 2y – sin xy + y 2 + c = 0

CH2_74 Example 3 Solve Solution: Rewrite the DE in the form (cos x sin x – xy 2 ) dx + y(1 – x 2 ) dy = 0 Since  M/  y = – 2xy =  N/  x (This DE is exact) Now  f/  y = y(1 – x 2 ) f(x, y) = ½y 2 (1 – x 2 ) + h(x)  f/  x = – xy 2 + h’(x) = cos x sin x – xy 2

CH2_75 Example 3 (2) We have h(x) = cos x sin x h(x) = -½ cos 2 x Thus ½y 2 (1 – x 2 ) – ½ cos 2 x = c 1 or y 2 (1 – x 2 ) – cos 2 x = c (7) where c = 2c 1. Now y(0) = 2, so c = 3. The solution is y 2 (1 – x 2 ) – cos 2 x = 3

CH2_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.

CH2_77  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  M y +  y M =  N x +  x N, or  x N –  y M = (M y – N x )  (9) Integrating Factors

CH2_78  Suppose  is a function of one variable, say x, then  x = d  /dx (9) becomes (10) If we have (M y – N x ) / 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 (N x – M y ) / M is a function of y only, then we can solve (11) for .

CH2_79  We summarize the results for M(x, y) dx + N(x, y) dy = 0(12) If (M y – N x ) / N depends only on x, then (13) If (N x – M y ) / M depends only on y, then (14)

CH2_80 Example 4 The nonlinear DE: xy dx + (2x 2 + 3y 2 – 20) dy = 0 is not exact. With M = xy, N = 2x 2 + 3y 2 – 20, we find M y = x, N x = 4x. Since depends on both x and y. depends only on y. The integrating factor is e  3dy/y = e 3lny = y 3 =  (y)

CH2_81 Example 4 (2) then the resulting equation is xy 4 dx + (2x 2 y 3 + 3y 5 – 20y 3 ) dy = 0 It is left to you to verify the solution is ½ x 2 y 4 + ½ y 6 – 5y 4 = c

CH2_ Solutions by Substitutions  Introduction If 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)).

CH2_83  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) = x 3 + y 3 is homogeneous of degree 3, f(tx, ty) = (tx) 3 + (ty) 3 = t 3 f(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) Homogeneous Equations

CH2_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), u=y/x(2) M(x, y)=y  M(v, 1), N(x, y)=y  N(v, 1), v=x/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 = 0 where u = y/x or y = ux and dy = udx + xdu,

CH2_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 = 0 or

CH2_86 Example 1 Solve (x 2 + y 2 ) dx + (x 2 – xy) dy = 0. Solution: We have M = x 2 + y 2, N = x 2 – xy are homogeneous of degree 2. Let y = ux, dy = u dx + x du, then (x 2 + u 2 x 2 ) dx + (x 2 - ux 2 )(u dx + x du) = 0

CH2_87 Example 1 (2) Then Simplify to get  Note: We may also try x = vy.

CH2_88  The DE: dy/dx + P(x)y = f(x)y n (4) where n is any real number, is called Bernoulli’s Equation.  Note for n = 0 and n = 1, (4) is linear, otherwise, let u = y 1-n to reduce (4) to a linear equation. Bernoulli’s Equation

CH2_89 Example 2 Solve x dy/dx + y = x 2 y 2. Solution: Rewrite the DE as dy/dx + (1/x)y = xy 2 With n = 2, then y = u -1, and dy/dx = -u -2 (du/dx) From the substitution and simplification, du/dx – (1/x)u = -x The integrating factor on (0,  ) is

CH2_90 Example 2 (2) Integrating gives x -1 u = -x + c, or u = -x 2 + cx. Since u = y -1, we have y = 1/u and a solution of the DE is y = 1/(−x 2 + cx).

CH2_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.

CH2_92 Example 3 Solve dy/dx = (-2x + y) 2 – 7, y(0) = 0. Solution: Let u = -2x + y, then du/dx = -2 + dy/dx, du/dx + 2 = u 2 – 7 or du/dx = u 2 – 9 This is separable. Using partial fractions, or

CH2_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.

CH2_94 Example 3 (3) The graph of the particular solution is shown in Fig 2.30 in solid color.

CH2_95 Fig 2.30

CH2_96  Using the Tangent Line Let us assume y’ = f(x, y), y(x 0 ) = y 0 (1) possess a solution. For example, the resulting graph is shown in Fig A Numerical Method

CH2_97 Fig 2.31

CH2_98  Using the linearization of the unknown solution y(x) of (1) at x 0, L(x) = f(x 0, y 0 )(x - x 0 ) + y 0 (2) Replacing x by x 1 = x 0 + h, we have L(x 1 ) = f(x 0, y 0 )(x 0 + h - x 0 ) + y 0 or y 1 = y 0 + h f(x 0, y 0 ) and y n+1 = y n + h f(x n, y n )(3) where x n = x 0 + nh. See Fig 2.32 Euler’s Method

CH2_99 Fig 2.32

CH2_100 Example 1 Consider Use Euler’s method to obtain y(2.5) using h = 0.1 and then h = Solution: Let the results step by step are shown in Table 2.1 and table 2.2.

CH2_101 Table 2.1 Table 2.2 Table 2.1 h = 0.1 xnxn ynyn Table 2.2 h = 0.05 xnxn ynyn

CH2_102 Example 2 Consider y’ = 0.2xy, y(1) = 1. Use Euler’s method to obtain y(1.5) using h = 0.1 and then h = Solution: We have f(x, y) = 0.2xy, the results step by step are shown in Table 2.3 and table 2.4.

CH2_103 Table 2.3 Table 2.3 h = 0.1 xnxn ynyn Actual Value Absolute Error % Rel. Error

CH2_104 Table 2.4 Table 2.4 h = 0.05 xnxn ynyn Actual Value Absolute Error % Rel. Error

CH2_105 Numerical Solver  See Fig 2.33 to know the comparisons of numerical methods. Fig 2.33

CH2_106  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. Using a Numerical Solver Fig 2.34

CH2_ Linear Models  Growth and Decay (1)

CH2_108 Example 1: Bacterial Growth P 0 : initial number of bacterial = P(0) P(1) = 3/2 P(0) Find the time necessary for triple number. Solution: Since dP/dt = kt, dP/dt – kt = 0, we have P(t) = ce kt, using P(0) = P 0 then c = P 0 and P(t) = P 0 e kt Since P(1) = 3/2 P(0), then P(1) = P 0 e k = 3/2 P(0) So, k = ln(3/2) = Now P(t) = P 0 e t = 3P 0, t = ln3/ = See Fig 2.35.

CH2_109 Fig 2.35

CH2_110 Fig 2.36  k > 0 is called a growth constant, and k > 0 is called a decay constant. See Fig 2.36.

CH2_111 Example 2: Half-Life of Plutonium A reactor converts U-238 into the isotope plutonium After 15 years, there is 0.043% of the initial amount A 0 of the plutonium has disintegrated. Find the half-life of this isotope. Solution: Let A(t) denote the amount of plutonium remaining at time t. The DE is as (2) The solution is A(t) = A 0 e kt. If 0.043% of A 0 has disintegrated, then % remains.

CH2_112 Example 2 (2) Then, A 0 = A(15) = A 0 e 15k, then k = (ln ) / 15 = Let A(t) = A 0 e t = ½ A 0 Then

CH2_113 Example 3: Carbon Dating A fossilized bone contains 1/1000 the original amount C-14. Determine the age of the fossil. Solution: We know the half-life of C-14 is 5600 years. Then A 0 /2 = A 0 e 5600k, k = −(ln 2)/5600 = − And A(t) = A 0 /1000 = A 0 e t

CH2_114  (3) where T m is the temperature of the medium around the object. Newton’s Law of Cooling

CH2_115 Example 4 A cake’s 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? Solution: We identify T m = 70, then (4) and T(3) = 200. From (4), we have

CH2_116 Example 4 (2) Using T(0) = 300 then c 2 = 230 Using T(3) = 200 then e 3k = 13/23, k = ThusT(t) = e t (5) From (5), only t = , T(t) = 70. It means we need a reasonably long time to get T = 70. See Fig 2.37.

CH2_117 Fig 2.37

CH2_118 Mixtures  (6)

CH2_119 Example 5 Recall from example 5 of Sec 1.3, we have How much salt is in the tank after along time? Solution: Since Using x(0) = 50, we have x(t) = e -t/100 (7) When t is large enough, x(t) = 600.

CH2_120 Fig 2.38

CH2_121 Series Circuits  See Fig (8) See Fig (9) (10)

CH2_122 Fig 2.39

CH2_123 Fig2.40

CH2_124 Example 6 Refer to Fig 2.39, where E(t) = 12 Volt, L = ½ Henry R = 10 Ohms. Determine i(t) where i(0) = 0. Solution: From (8), Then Using i(0) = 0, c = -6/5, then i(t) = (6/5) – (6/5)e -20t.

CH2_125 Example 6 (2) A general solution of (8) is (11) When E(t) = E 0 is a constant, (11) becomes (12) where the first term is called a steady-state part, and the second term is a transient term.

CH2_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.

CH2_127 Fig 2.41

CH2_ Nonlinear Models  Population Dynamics If 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.

CH2_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.

CH2_130 Fig 2.46

CH2_131  Suppose f (P) = c 1 P + c 2. Using the conditions, we have c 2 = r, c 1 = −r/K. Then (2) becomes (3) Relabel (3), then (4) which is known as a logistic equation, its solution is called the logistic function and its graph is called a logistic curve.

CH2_132 Solution of the Logistic Equation  From After simplification, we have

CH2_133  If P(0) = P 0  a/b, then c 1 = P 0 /(a – bP 0 ) (5)

CH2_134 Graph of P(t)  Form (5), we have the graph as in Fig When 0 < P 0 < a/2b, see Fig 2.47(a). When a/2b < P 0 < a/b, see Fig 2.47(b). Fig 2.47

CH2_135 Example 1 Form the previous discussion, assume an isolated campus of 1000 students, then we have the DE Determine x(6). Solution: Identify a = 1000k, b = k, from (5)

CH2_136 Example 1 (2) Since x(4) = 50, then -1000k = , Thus x(t) = 1000/( e t ) See Fig 2.48.

CH2_137 Fig 2.48

CH2_138 Modification of the Logistic Equation  or (6) or (7) which is known as the Gompertz DE.

CH2_139 Chemical Reactions  (8) or (9)

CH2_140 Example 2 The chemical reaction is described as Then By separation of variables and partial fractions, (10) Using X(10) = 30, 210k = , finally (11) See Fig 2.49.

CH2_141 Fig 2.49

CH2_ Modeling with Systems of First-Order DEs  Systems (1) where g 1 and g 2 are linear in x and y.  Radioactive Decay Series (2)

CH2_143  From Fig 2.52, we have (3) Mixtures

CH2_144 Fig 2.52

CH2_145  Let x, y denote the fox and rabbit populations at t. When lacking of food, dx/dt = – ax, a > 0(4) When rabbits are present, dx/dt = – ax + bxy(5) When lacking of foxes, dy/dt = dy, d > 0(6) When foxes are present, dy/dt = dy – cxy(7) A Predator-Prey Model

CH2_146 Then (8) which is known as the Lotka-Volterra predator-prey model.

CH2_147 Example 1  Suppose Figure 2.53 shows the graph of the solution.

CH2_148 Fig 2.53

CH2_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 = a 1 x – b 1 x 2 dy/dt = a 2 y – b 2 y 2 (12)

CH2_150 or dx/dt = a 1 x – b 1 x 2 – c 1 xy dy/dt = a 2 y – b 2 y 2 – c 2 xy(13)

CH2_151 Network  Referring to Fig 2.54, we have i 1 (t) = i 2 (t) + i 3 (t)(14) (15) (16)

CH2_152  Using (14) to eliminate i 1, then (17) Referring to Fig 2.55, please verify (18)

CH2_153 Fig 2.54

CH2_154 Fig 2.55