First-Order Differential Equations CHAPTER 2 First-Order Differential Equations

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

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

Fig2.1

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)

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).

Fig2.2

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.

Fig2.3

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.

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.

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.

Fig2.4

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 < 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).

Fig 2.5

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.

(4)If y(x) is bounded above by c1, (y(x) < c1), the graph of y(x) will approach y(x) = c1; If c1 < y(x) < c2, it will approach y(x) = c1 and y(x) = c2; If c2 < y(x) , it will approach y(x) = c2;

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.

Fig 2.6

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.

Fig2.7

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.

Fig 2.8

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.

Fig 2.9

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 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

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

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

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).

Example 2 Solve Solution: We also can rewrite the solution as x2 + y2 = c2, where c2 = 2c1 Apply the initial condition, 16 + 9 = 25 = c2 See Fig2.18.

Fig2.18

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.

Example 3 Solve dy/dx = y2 – 4. Solution: Rewrite this DE as (5) then

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.

Example 4 Solve Solution: Rewrite 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)

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.

Fig2.19 Fig2.20

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

Fig2.21

2.3 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 a1(x)(dy/dx) + a0(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

Standard Form The standard for of a first-order DE can be written as 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 = 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).

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

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

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)

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

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

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-3xy = -2e-3x + c, a solution is y = -2 + ce-3x, - < x < .

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

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)

Example 2 Solve Solution: Dividing both sides by x, we have (10) So, P(x) = –4/x, f(x) = x5ex, P and f are continuous on (0, ). Since x > 0, we write the integrating factor as

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

Example 3 Find the general solution of Solution: Rewrite 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

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.

Example 4 Solve Solution: We 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-x Since y(0) = 4, then c = 5. The solution is y = x – 1 + 5e-x, – < x <  (12)

Notes: From the above example, we find 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

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

Example 5 (2) We solve this problem on 0  x  1 and 1 < x < . For 0  x  1, then y = 1 + c1e-x Since y(0) = 0, c1 = -1, y = 1 - e-x For x > 1, dy/dx + y = 0 then y = c2e-x

Example 5 We have Furthermore, 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, ).

Fig2.26

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

Example 6 Solve dy/dx – 2xy = 2, y(0) = 1. Solution: We find the integrating factor is exp{-x2}, we get (15) Applying y(0) = 1, we have c = 1. See Fig2.27

Fig2.27

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.

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)

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

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

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)

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)

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.

Example 1 Solve 2xy dx + (x2 – 1) dy = 0. Solution: With M(x, y) = 2xy, N(x, y) = x2 – 1, we have M/y = 2x = N/x Thus it is exact. There exists a function f such that f/x = 2xy, f/y = x2 – 1 Then f(x, y) = x2y + g(y) f/y = x2 + g’(y) = x2 – 1 g’(y) = -1, g(y) = -y

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.

Example 2 Solve (e2y – y cos xy)dx+(2xe2y – x cos xy + 2y)dy = 0. Solution: This DE is exact because M/y = 2e2y + xy sin xy – cos xy = N/x Hence a function f exists, and f/y = 2xe2y – x cos xy + 2y that is,

Example 2 (2) Thus h’(x) = 0, h(x) = c. The solution is xe2y – sin xy + y2 + c = 0

Example 3 Solve Solution: Rewrite 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

Example 3 (2) We have h(x) = cos x sin x h(x) = -½ cos2 x Thus 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

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.

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)

Suppose  is a function of one variable, say x, then 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 .

We summarize the results for. M(x, y) dx + N(x, y) dy = 0 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)

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)

Example 4 (2) then the resulting equation is xy4 dx + (2x2y3 + 3y5 – 20y3) dy = 0 It is left to you to verify the solution is ½ x2y4 + ½ y6 – 5y4 = c

2.5 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)).

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)

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,

then. M(1, u) dx + N(1, u)(u dx + x du) = 0, and 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

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

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

Bernoulli’s Equation The DE: dy/dx + P(x)y = f(x)yn (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 = y1-n to reduce (4) to a linear equation.

Example 2 Solve x dy/dx + y = x2y2. Solution: Rewrite the DE as dy/dx + (1/x)y = xy2 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

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).

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.

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 = u2 – 7 or du/dx = u2 – 9 This is separable. Using partial fractions, or

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.

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

Fig 2.30

2.6 A Numerical Method Using the Tangent Line Let us assume y’ = f(x, y), y(x0) = y0 (1) possess a solution. For example, the resulting graph is shown in Fig 2.31.

Fig 2.31

Euler’s 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 yn+1 = yn + h f(xn, yn) (3) where xn = x0 + nh. See Fig 2.32

Fig 2.32

Example 1 Consider Use Euler’s method to obtain y(2.5) using h = 0.1 and then h = 0.05. Solution: Let the results step by step are shown in Table 2.1 and table 2.2.

Table 2.1 Table 2.2 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 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

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 = 0.05. Solution: We have f(x, y) = 0.2xy, the results step by step are shown in Table 2.3 and table 2.4.

Table 2.3 Table 2.3 h = 0.1 xn yn Actual Value Absolute Error % Rel. Error 1.00 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

Table 2.4 Table 2.4 h = 0.05 xn yn Actual Value Absolute Error % Rel. Error 1.00 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

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

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

2.7 Linear Models Growth and Decay (1)

Example 1: Bacterial Growth P0 : 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) = cekt, using P(0) = P0 then 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.

Fig 2.35

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

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. Solution: Let 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.

Example 2 (2) Then, 0.99957A0 = A(15) = A0e15k, then k = (ln 0.99957) / 15 =-0.00002867 Let A(t) = A0e-0.00002867t = ½ A0 Then

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 A0 /2 = A0e5600k, k = −(ln 2)/5600 = −0.00012378. And A(t) = A0 /1000 = A0e -0.00012378t

Newton’s Law of Cooling (3) where Tm is the temperature of the medium around the object.

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 Tm = 70, then (4) and T(3) = 200. From (4), we have

Example 4 (2) Using T(0) = 300 then c2 = 230 Using T(3) = 200 then e3k = 13/23, k = -0.19018 Thus 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.

Fig 2.37

Mixtures (6)

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) = 600 - 550e-t/100 (7) When t is large enough, x(t) = 600.

Fig 2.38

Series Circuits See Fig 2.39. (8) See Fig 2.40. (9) (10)

Fig 2.39

Fig2.40

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.

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.

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.

Fig 2.41

2.8 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.

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.

Fig 2.46

Suppose f (P) = c1P + c2. Using the conditions, we have c2 = r, c1 = −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.

Solution of the Logistic Equation From After simplification, we have

If P(0) = P0  a/b, then c1 = P0/(a – bP0) (5)

Graph of P(t) Form (5), we have the graph as in Fig 2.47. When 0 < P0 < a/2b, see Fig 2.47(a). When a/2b < P0 < a/b, see Fig 2.47(b). Fig 2.47

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)

Example 1 (2) Since x(4) = 50, then -1000k = -0.9906, Thus x(t) = 1000/(1 + 999e-0.9906t) See Fig 2.48.

Fig 2.48

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

Chemical Reactions (8) or (9)

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.

Fig 2.49

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)

Mixtures From Fig 2.52, we have (3)

Fig 2.52

A Predator-Prey Model 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)

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

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

Fig 2.53

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)

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

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

Using (14) to eliminate i1, then. (17) Referring to Fig 2 Using (14) to eliminate i1, then (17) Referring to Fig 2.55, please verify (18)

Fig 2.54

Fig 2.55

Thank You !