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First-Order Differential Equations

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### 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.
- 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

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 DEdy/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).

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.

- Increasing/DecreasingIf 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.

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, R2and R3as in Fig 2.5(b).

- 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) = c1and y(x) = c2;If c2 < y(x), it will approach y(x) = c2;

Example 4

Referring to example 3, P =0and 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) = P0and when a solution pass through P0,wehave three kind of graph according to the interval where P0 lies on. See Fig 2.6.

Example 5

The DE: dy/dx = (y – 1)2possesses 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.

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.

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.

Separable Equations

A first-order DE of the formdy/dx = g(x)h(y)is said to be separable or to have separable variables.

2.2 Separable Variables- Introduction: Consider dy/dx = f(x, y) = g(x).The DEdy/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

- 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(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 haveReplacing by c, gives y = c(1 + x).

Example 2

Solve

Solution:We also can rewrite the solution asx2+ y2 = c2,where c2 =2c1Apply the initial condition, 16 + 9 = 25 = c2See 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 (2)

Replacing exp(c2)by c andsolving for y, we have(6)If we rewrite the DE as dy/dx = (y + 2)(y – 2),from the previous discussion, we have y = 2is a singular solution.

Example 4

Solve

Solution:Rewrite this DE asusing sin2x = 2sin x cos x, then (ey – ye-y) dy = 2sin x dx from integration by parts,ey + ye-y + e-y = -2cos x + c (7)From y(0) =0,we have c = 4 to getey + 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.

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

Linear Equations

A first-order DE of the forma1(x)(dy/dx) + a0(x)y = g(x) (1)is said to be a linear equation iny. When g(x) = 0,(1) is said to be homogeneous; otherwise it is nonhomogeneous.

2.3 Linear Equations- Introduction:Linear DEs are friendly to be solved. We can find some smooth methods to deal with.

- Standard FormThe standard for of a first-order DE can be written asdy/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 equationdy/dx + P(x)y = 0 (3)and yp is a particular solution of (2).

- VerificationNow (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 ypinto (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 = –2is 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 Pandfare continuous onI. 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, ) isx-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 , sogives exy = xex – ex + c and y = x – 1 + ce-xSince y(0) = 4,then c = 5. The solution isy = x – 1+ 5e-x, – < x < (12)

- Notes: From the above example, we find yc = ce-xand yp = x – 1we call ycis a transient term, since yc 0as x .Some solutions are shown in Fig2.24.

Fig2.24

Example 5 (2)

We solve this problem on 0 x 1 and 1 <x < .For 0 x 1, then y = 1 + c1e-xSince y(0)= 0, c1 = -1, y = 1 - e-xFor x > 1, dy/dx + y = 0then y = c2e-x

Example 5

We haveFurthermore, we want y(x)iscontinuous 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, ).

Functions Defined by Integrals

- We are interested in the error function and complementary error functionand(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

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 differentialor 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)

Exact Equation

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 = 0is said to be an exact equation, if the left side is an

exact differential.

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)

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 RM(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)

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/xThus it is exact. There exists a function f such thatf/x =2xy, f/y =x2 – 1Thenf(x, y) = x2y + g(y)f/y = x2 + g’(y) =x2 – 1g’(y)=-1, g(y) = -y

Example 1 (2)

Hence f(x, y) = x2y – y, and the solution isx2y – 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 becauseM/y = 2e2y + xy sin xy – cos xy = N/xHence a function f exists, and f/y = 2xe2y – x cos xy + 2ythat is,

Example 2 (2)

Thus h’(x) = 0, h(x) = c. The solution isxe2y – 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 = 0SinceM/y = – 2xy = N/x (This DE is exact)Nowf/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) =-½cos2 xThus ½y2(1– x2) –½cos2 x = c1or y2(1 – x2) –cos2x = c (7)where c = 2c1. Now y(0) = 2, so c = 3.The solution isy2(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, orxN– yM = (My – Nx) (9)

- 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 forM(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 = 0is not exact. With M = xy, N =2x2 + 3y2 – 20, we find My = x, Nx = 4x. Sincedepends 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 isxy4dx + (2x2y3 +3y5– 20y3) dy = 0It is left to you to verify the solution is ½ x2y4 + ½ y6 – 5y4 = c

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, thenSince 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) = tf(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) = tM(x, y), N(tx, ty) = tN(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)=xN(1, u), u=y/x (2)M(x, y)=y M(v, 1), N(x, y)=yN(v, 1), v=x/y (3)Then (1) becomesx M(1, u) dx + x N(1, u) dy = 0,orM(1, u) dx + N(1, u) dy = 0where u = y/x or y = ux and dy = udx + xdu,

then Sec 2.3. M(1, u) dx + N(1, u)(u dx + x du) = 0,and[M(1, u) + uN(1, u)] dx + xN(1, u) du = 0or

Example 1 Sec 2.3.

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

Bernoulli’s Equation Sec 2.3.

- 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, letu = y1-n to reduce (4) to a linear equation.

Example 2 Sec 2.3.

Solve x dy/dx + y = x2y2.

Solution:Rewrite the DE as dy/dx + (1/x)y = xy2With n = 2, then y = u-1, anddy/dx = -u-2(du/dx) From the substitution and simplification, du/dx – (1/x)u = -xThe integrating factor on (0, ) is

Example 2 (2) Sec 2.3.

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 Sec 2.3.

- 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 Sec 2.3.

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

Example 3 (2) Sec 2.3.

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

Example 3 (3) Sec 2.3.

The graph of the particular solutionis shown in Fig 2.30 in solid color.

Fig 2.30 Sec 2.3.

2.6 A Numerical Method Sec 2.3.

- 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.

Fig 2.31 Sec 2.3.

Euler’s Method Sec 2.3.

- 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 haveL(x1) = f(x0, y0)(x0 + h - x0) + y0or 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 Sec 2.3.

Example 1 Sec 2.3.

Consider Use Euler’s method to obtain y(2.5) using h = 0.1andthen 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 Sec 2.3.

Example 2 Sec 2.3.

Consider y’ =0.2xy, y(1)=1. Use Euler’s method to obtain y(1.5)using h = 0.1andthen 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 Sec 2.3.

Table 2.4 Sec 2.3.

Using a Numerical Solver Sec 2.3.

- 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 Sec 2.3.

- Growth and Decay (1)

Example 1: Bacterial Growth Sec 2.3.

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) = P0then c = P0 and P(t) = P0ektSince 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 Sec 2.3.

Fig 2.36 Sec 2.3.

- 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 Sec 2.3.

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) Sec 2.3.

Then, 0.99957A0 = A(15) = A0e15k, then k = (ln0.99957) / 15=-0.00002867Let A(t) = A0e-0.00002867t= ½ A0Then

Example 3: Carbon Dating Sec 2.3.

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 Sec 2.3.

- (3)where Tmisthe temperature of the medium around the object.

Example 4 Sec 2.3.

A cake’s temperature is 300F. Three minutes later its temperature is 200F. How long will it for this cake to cool off to a room temperature of 70F?

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

Example 4 (2) Sec 2.3.

Using T(0) = 300then c2 = 230Using T(3) = 200then 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.

Fig 2.37 Sec 2.3.

Mixtures Sec 2.3.

- (6)

Example 5 Sec 2.3.

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

Fig 2.38 Sec 2.3.

Series Circuits Sec 2.3.

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

Fig 2.39 Sec 2.3.

Fig2.40 Sec 2.3.

Example 6 Sec 2.3.

Refer to Fig 2.39, where E(t) = 12 Volt, L = ½ HenryR = 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) Sec 2.3.

A general solution of (8) is (11)When E(t) = E0is a constant, (11) becomes (12)where the first term is called a steady-state part, and the second term is a transient term.

Note: Sec 2.3.

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 Sec 2.3.

2.8 Nonlinear Models Sec 2.3.

- Population DynamicsIf P(t) denotes the size of population at t, the relative(orspecific), 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 Sec 2.3.

- 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 Sec 2.3.

- Suppose Sec 2.3. 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 Sec 2.3.

- From After simplification, we have

- If Sec 2.3. P(0) = P0 a/b, then c1 = P0/(a – bP0)(5)

Graph of Sec 2.3. 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 Sec 2.3.

Form the previous discussion, assume an isolated campus of 1000 students, then we have the DEDetermine x(6).

Solution:Identify a = 1000k, b = k, from (5)

Example 1 (2) Sec 2.3.

Since x(4)= 50, then -1000k = -0.9906,Thusx(t) = 1000/(1 +999e-0.9906t)

See Fig 2.48.

Fig 2.48 Sec 2.3.

Modification of the Logistic Equation Sec 2.3.

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

Chemical Reactions Sec 2.3.

- (8)or(9)

Example 2 Sec 2.3.

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

Fig 2.49 Sec 2.3.

2.9 Modeling with Systems of First-Order DEs Sec 2.3.

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

Mixtures Sec 2.3.

- From Fig 2.52, we have(3)

Fig 2.52 Sec 2.3.

A Predator-Prey Model Sec 2.3.

- 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 Sec 2.3. (8)which is known as the Lotka-Volterra predator-prey model.

Example 1 Sec 2.3.

- Suppose Figure 2.53 shows the graph of the solution.

Fig 2.53 Sec 2.3.

Competition Models Sec 2.3.

- 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 – b1x2dy/dt = a2y – b2y2(12)

or Sec 2.3. dx/dt = a1x – b1x2 – c1xydy/dt = a2y – b2y2 – c2xy(13)

Network Sec 2.3.

- Referring to Fig 2.54, we havei1(t) = i2(t) + i3(t)(14)(15)(16)

- Using (14) to eliminate Sec 2.3. i1, then(17)Referring to Fig 2.55, please verify (18)

Fig 2.54 Sec 2.3.

Fig 2.55 Sec 2.3.

Thank You ! Sec 2.3.

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