First-Order Differential Equations

CHAPTER 2

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Contents

2.1 Solution Curves Without a Solution2.2 Separable Variables2.3 Linear Equations2.4 Exact Equations2.5 Solutions by Substitutions2.6 A Numerical Methods2.7 Linear Models2.8 Nonlinear Model2.9 Modeling with Systems of First-Order DEs

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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

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Fig2.1

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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)

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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).

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Fig2.2

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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.

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Fig2.3

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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.

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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.

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Example 3

The following DEdP/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.

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Fig2.4

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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).

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Fig 2.5

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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.

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(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;

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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.

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Fig 2.6

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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.

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Fig2.7

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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.

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Fig 2.8

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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.

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Fig 2.9

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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

A first-order DE of the formdy/dx = g(x)h(y)

is said to be separable or to have separable variables.

DEFINITION 2.1Separable Equations

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Rewrite the above equation as

(2)

where p(y) = 1/h(y). When h(y) = 1, (2) reduces to (1).

)()( xgdxdy

yp

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If y = (x) is a solution of (2), we must have

and

(3)

But dy = (x) dx, (3) is the same as

(4)

dxxxP )('))((

)()('))(( xgxxP

cxGyHdxxgdyyP )()(or )()(

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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).

)1(

1

1lnln

1

1

111 1ln1ln

1

xe

exeeey

cxy

xdx

ydy

c

ccxcx

1ce

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Example 2

Solve

Solution:

We also can rewrite the solution asx2 + y2 = c2, where c2 = 2c1

Apply the initial condition, 16 + 9 = 25 = c2

See Fig2.18.

3)4( , yyx

dxdy

1

22

22 and c

xyxdxydy

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Fig2.18

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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.

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Example 3

Solve dy/dx = y2 – 4.

Solution:Rewrite this DE as

(5)then

dxdyyy

dxy

dy

22or

44

1

4

1

2

22

,422

ln

,241

2ln41

24

2

1

cxeyy

cxyy

cxyy

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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.

x

x

ce

cey

4

4

1

12

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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 getey + ye-y + e-y = 4 −2 cos x (8)

0)0( ,2sin)(cos 2 yxedxdy

yex yy

dxxx

dye

yey

y

cos2sin2

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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.

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Fig2.19 Fig2.20

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If we solve dy/dx = xy½ , y(0) = 0

(9)The resulting graphs are shown in Fig2.21.

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Fig2.21

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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 forma1(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 2.2Linear Equations

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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).

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Verification

Now (3) is also separable. Rewrite (3) as

Solving for y gives

)()()(

])[(][

xfyxPdx

dyyxP

dxdy

yyxPyydxd

Pp

cc

pcoc

0)( dxxPy

dy

)(1)(

xcyceydxxP

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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

)()(

or

)()(

111

111

xfdxdu

yyxPdxdy

u

xfuyxPdxdu

ydxdy

u

Variation of Parameters

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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)

dxxyxf

udxxyxf

du)()(

and )()(

11

dxxfeeceyyydxxPdxxPdxxP

pc

)()()()(

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Solving Procedures

If (4) is multiplied by(5)

then(6)

is differentiated(7)

we get(8)

Dividing (8) by gives (2).

dxxPe

)(

dxxfecye dxxPdxxP )()()(

)()()(

xfeyedxd dxxPdxxP

)()()()()(

xfeyexPdxdy

edxxPdxxPdxxP

dxxPe )(

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

dxxPe )(

Integrating Factor

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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 < .

xdxee 3)3(

xxx eyedxdy

e 333 63

xx eyedxd 33 6][

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The DE of example 1 can be written as

so that y = –2 is a critical point.

)2(3 ydxdy

Notes

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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) 00)( ),()( yxyxfyxPdxdy

General Solutions

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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

xexydxdy

x 64

xexyxdx

dy 54

4lnln4/4 4

xeee xxxdx

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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

xxy xeyxdxd

xexdxdy

x ][ ,4 454

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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

092

y

x

xdxdy

929ln2

1)9/(2

2

1)9/(

222

xeee

xxxdxxxdx

0)9( 2 xydxdy

x

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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.

cyx 92

092 yxdxd

9/ 2 xcy

9/ 2 xcy

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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 isy = x – 1 + 5e-x, – < x < (1

2)

4)0( , yxydxdy

xxdxee /

xx xeyedxd ][

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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

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Example 5

Solve , where

Solution:First we see the graph of f(x) in Fig2.25.

Fig2.25

1 ,0

10 ,1)(

x

xxf

0)0( ),( yxfydxdy

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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

xx eyedxd

ydxdy ][ ,1

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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, ).

1 ,)1(

10 ,1

xee

xey

x

x

1 ,

10 ,1

2 xec

xey

x

x

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Fig2.26

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We are interested in the error function and complementary error function

and (14)

Since , we find erf(x) + erfc(x) = 1

dtexx t

0

22)(erf

dtex

x

t

22)(erfc

1)/2(0

2

dte t

Functions Defined by Integrals

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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

22

2][ xx eedxd

222

02 xx tx cedteey

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Fig2.27

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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.

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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)

dyyf

dxxf

dz

0

dyyf

dxxf

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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 2.3Exact Equation

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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 2.1Criterion for an Extra Differential

xN

yM

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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)

xN

yf

xxyf

xf

yyM

2

xf

yf

xf

yf

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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)

dxyxMy

yxNxg ) ,() ,()('

) ,()(') ,( yxNygdxyxMyy

f

)() () ,( ygdxyx,Myxf

Method of Solution

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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.

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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 that

f/x = 2xy, f/y = x2 – 1Then

f(x, y) = x2y + g(y)f/y = x2 + g’(y) = x2 – 1g’(y) = -1, g(y) = -y

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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.

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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/xHence a function f exists, and

f/y = 2xe2y – x cos xy + 2ythat is,

xyyexhxyyexf

xhyxyxe

ydyxydyxdyexyxf

yy

y

y

cos)('cos

)(sin

2cos2) ,(

22

22

2

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Example 2 (2)

Thus h’(x) = 0, h(x) = c. The solution isxe2y – sin xy + y2 + c = 0

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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

2)0( ,)1(

sincos2

2

yxy

xxxydxdy

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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

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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.

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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)

Integrating Factors

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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 .

N

NM

dyd xy

M

MN

dyd yx

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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)

dxN

NM xy

ex

)(

dyM

MN yx

ey

)(

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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)

2032

3

2032

42222

yx

x

yx

xxN

NM xy

yM

MN yx 3

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Example 4 (2)

then the resulting equation isxy4 dx + (2x2y3 + 3y5 – 20y3) dy = 0

It is left to you to verify the solution is½ x2y4 + ½ y6 – 5y4 = c

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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)).

dxdu

uxguxgdxdy

ux ),() ,(

dxdu

uxguxguxgxf ux ) ,() ,()) ,( ,(

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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)

Homogeneous Equations

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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) 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,

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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

0) ,1() ,1(

) ,1(

uuNuM

dxuNx

dx

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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

011

xdx

duuu

01

21

xdx

duu

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Example 1 (2)

Then

Simplify to get

Note: We may also try x = vy.

cxxy

xy

cxuu

lnln1ln2

lnln1ln2

xycxeyxxy

csyx /2

2

)(or )(

ln

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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.

Bernoulli’s Equation

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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, anddy/dx = -u-2(du/dx)

From the substitution and simplification, du/dx – (1/x)u = -x

The integrating factor on (0, ) is

1lnln/ 1

xeee xxxdx

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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).

xuxdx

du 1

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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.

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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 – 9This is separable. Using partial fractions,

or

dxuu

du )3)(3(

dxduuu

31

31

61

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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.

133

ln61

cxuu

x

x

ce

cexy 6

6

1

)1(32

x

x

ce

ceu 6

6

1

)1(3

x

x

e

exuxy 6

6

1

)1(322

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Example 3 (3)

The graph of the particular solution

is shown in Fig 2.30 in solid color.

x

x

e

exuxy 6

6

1

)1(322

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Fig 2.30

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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.

2.6 A Numerical Method

4)2( ,4.01.0' 2 yxyy

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Fig 2.31

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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

Euler’s Method

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Fig 2.32

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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.

.4)2( ,4.01.0' 2 yxyy

24.01.0) ,( xyyxf

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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

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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.

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Table 2.3

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

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Table 2.4

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

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Numerical Solver

See Fig 2.33 to know the comparisons of numerical methods.

Fig 2.33

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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

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2.7 Linear Models

Growth and Decay

(1)00)( , xtxkx

dtdx

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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.

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Fig 2.35

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Fig 2.36

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

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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.

0)0( , AAkAdtdA

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Example 2 (2)

Then, 0.99957A0 = A(15) = A0e15k, thenk = (ln 0.99957) / 15 =-0.00002867

Let A(t) = A0e-0.00002867t = ½ A0

Then years. 24180

00002867.02ln T

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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

years. 558000.00012378

1000ln T

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(3)

where Tm is the temperature of the medium around the object.

)( mTTkdxdT

Newton’s Law of Cooling

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Example 4

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

300)0( ),70( TTkdxdT

ktectT

cktTkdtT

dT

2

1

70)(

70ln ,70

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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.

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Fig 2.37

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Mixtures

(6)outin RR

dtdx

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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.

50)0( ,6100

1 xxdtdx

100/100/100/ 600)( ,6][ ttt cetxexedtd

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Fig 2.38

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Series Circuits

See Fig 2.39. (8)

See Fig 2.40.

(9)

(10)

)(tERidtdi

L

)(1

tEqC

Ri

)(1

tEqCdt

dqR

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Fig 2.39

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Fig2.40

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Example 6

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.

0)0( ,121021 ii

dtdi

t

t

ceti

eiedtd

20

2020

56

)(

24][

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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.

tLRtLRtLR

cedttEeL

eti )/()/(

)/(

)()(

tLRo ceRE

ti )/()(

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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.

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Fig 2.41

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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.

PdtdP /

)(or )(/

PPfdtdP

PfP

dtdP

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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.

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Fig 2.46

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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.

P

Kr

rPdtdP

)( bPaPdtdP

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Solution of the Logistic Equation

From

After simplification, we have

dtbPaP

dP )(

acatPba

P

dtdPbPa

b/aP

a

)(ln

lnln

1

atat

at

ebc

ac

ebc

eactP

1

1

1

1

1)(

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

(5) atebPabP

aPtP

)(

)(00

0

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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

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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)

1)0( ,)1000( xxkxdtdx

ktetx 10009991

1000)(

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Example 1 (2)

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

See Fig 2.48.

students 2769991

1000)6( 9436.5

e

x

tetx 9906.09991

1000)(

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Fig 2.48

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Modification of the Logistic Equation

or

(6)

or

(7)

which is known as the Gompertz DE.

hbPaPdtdP )(

hbPaPdtdP )(

)ln( PbaPdtdP

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Chemical Reactions

(8)

or

(9)

X

NMN

bXNM

Ma

dtdX

))(( XXkdtdX

CH2_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.

X

XdtdX

54

325

50

)40)(250( XXkdtdX

ktecXX

cktXX 210

21 40250

or 21040250

ln

t

t

e

etX 1258.0

1258.0

425

11000)(

CH2_141

Fig 2.49

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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)

) , ,(1 yxtgdtdx ) , ,(2 yxtg

dtdy

ydtdz

yxdtdy

xdtdx

2

21

1

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From Fig 2.52, we have

(3)21

2

211

252

252

501

252

xxdt

dx

xxdtdx

Mixtures

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Fig 2.52

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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

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Then

(8)

which is known as the Lotka-Volterra predator-prey model.

)(

)(

cxdycxydydtdy

byaxbxyaxdtdx

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Example 1

Suppose

Figure 2.53 shows the graph of the solution.

4)0( ,4)0( ,9.05.4

08.016.0

yxxyydtdy

xyxdtdx

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Fig 2.53

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Competition Models

dx/dt = ax, dy/dt = cy (9)Two species compete, then

dx/dt = ax – bydy/dt = cy – dx (10)

or dx/dt = ax – bxydy/dt = cy – dxy (11)

or dx/dt = a1x – b1x2

dy/dt = a2y – b2y2 (12)

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ordx/dt = a1x – b1x2 – c1xydy/dt = a2y – b2y2 – c2xy (1

3)

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Network

Referring to Fig 2.54, we havei1(t) = i2(t) + i3(t)

(14)

(15)

(16)

dtdi

LRitE

Ridtdi

LRitE

3211

222

111

)(

)(

CH2_152

Using (14) to eliminate i1, then

(17)

Referring to Fig 2.55, please verify

(18)

)(

)()(

31213

2

332212

1

tEiRiRdtdi

L

tEiRiRRdtdi

L

0

)(

122

21

iidtdi

RC

tERidtdi

L

CH2_153

Fig 2.54

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Fig 2.55