Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • MTH16_Lec-14_sec_9-4_ODE_SlopeFields_Euler.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics§9.4 ODE

Analytics



Review §

Any QUESTIONS About• §9.3 Differential Equation Applications

Any QUESTIONS About HomeWork• §9.3 → HW-15

9.3



§9.4 Learning Goals Analyze solutions of

differential equationsusing slope fields

Use Euler’s methodfor approximating solutions of initialvalue problems



Slope Fields Recall that indefinite integration, or

AntiDifferentiation, is the process of reverting a function from its derivative. • In other words, if we have a derivative, the

AntiDerivative allows us to regain the function before it was differentiated – EXCEPT for the CONSTANT, of course.

Given the derivative dy/dx = f ‘(x) then solving for y (or f(x)), produces the General Solution of a Differential Eqn



Slope Fields AntiDifferentiation (Separate Variables)

Example• Let: • Then Separating the Variables:• Now take the AntiDerivative: • To Produce the General Solution:

This Method Produces an EXACT and SYMBOLIC Solution which is also called an ANALYTICAL Solution

xdxdy 2

dxxdy 2

dxxdy 2

Cxy 2



Slope Fields Slope Fields, on the other hand,

provide a Graphical Method for ODE Solution

Slope, or Direction, fields basically draw slopes at various CoOrdinates for differing values of C.

Example: The Slope Field for ODE x

dxdy



Slope Fields slope field describes

several different parabolas based on varying values of C

Slope Field Example: create the slope field for the Ordinary Differential Eequation:

Cxydxxdyxdxdy

21

2

yx

dxdy



Slope Fields Note that dy/dx = x/y calculates the

slope at any (x,y) CoOrdinate point• At (x,y) = (−2, 2),

dy/dx = −2/2 = −1• At (x,y) = (−2, 1),

dy/dx = −2/1 = −2• At (x,y) = (−2, 0),

dy/dx = −2/0 = UnDef.• And SoOn

Produces OutLine of a HYPERBOLA

x

y

-2

-1

1

2

-2 -1 1 2

x

y



Slope Fields Of course this Variable Separable ODE

can be easily solved analytically

yx

dxdy

dxxdyy dxxdyy

Cxy 22

21

21

Cxy 22

Cyx 22

x

y

-2

-1

1

2

-2 -1 1 2

x

y



Slope Fields Example For the given slope field, sketch two

approximate solutions – one of which is passes through(4,2):• Solve ODE Analytically using

using (4,2) BC

12 xm

2,41

21

xdxdy dxxdy

1

21

dxxdy

1

21

C 44412 2

Cxxy 2

41

C 2 241 2 xxySoln



Slope Field Identification

C

3xdxdy

In order to determine a slope field from a

differential equation, we should consider the

following:i) If isoclines (points with the

same slope) are along horizontal lines, then DE depends only on y

ii) Do you know a slope at a particular point?

iii) If we have the same slope along vertical lines, then DE depends only on x

iv) Is the slope field sinusoidal?v) What x and y values make the

slope 0, 1, or undefined?vi) dy/dx = a(x ± y) has similar

slopes along a diagonal.vii) Can you solve the separable

DE?

1. _____

2. _____

3. _____

4. _____

5. _____

6. _____

7. _____

8. _____

Match the correct DE with its graph:2y

dxdy

xdxdy cos

xdxdy sin

yxdxdy

22 yxdxdy

1 yydxdy

yx

dxdy

A B

C

E

G

D

F

H

H

B

F

D

G

E

A



Example Demand Slope Field Imagine that the change in

fraction of a production facility’s inventory that is demanded, D, each period is given by• Where p is the unit price in $k

Draw a slope field to approximate a solution assuming a half-stocked (50%) inventory and $2k per item, and then • Verify the Slope-Field solution using

Separation of Variables. c



Example Demand Slope Field SOLUTION: Calculate some

Slope Values from peDm

dpdD 1

1100,0 0 emdpdD

0111,1 1 emdpdD

068.015.05.0,2 2 emdpdD



Example Demand Slope Field An approximate

solution passing through (2,0.5) with slope field on the window 0 < x < 3 and 0 < y < 1

$k/unit p

fr

actio

nal

p

D



Example Demand Slope Field Find an exact solution to this differential

equation using separation of variables:

Remove absolute-value and then change signs as inventory demanded satisfies: 0≤ D ≤1



Example Demand Slope Field Removing ABS Bars

Or Now use Boundary Value ($2k/unit,0.5)

CeCeDp eeDeeCeDpp

11ln 1ln

pp eeC eADDeeD 110with 1

572.05.02

eeA



Example Demand Slope Field Graph for

This is VERY SIMILAR to the Slope Field Graph Sketched Before

peepD 572.01



Numerical ODE Solutions Next We’ll “look

under the hood” of NUMERICAL Solutions to ODE’s

The BASIC Game-Plan for even the most Sophisticated Solvers:• Given a STARTING

POINT, y(0)• Use ODE to find dy/dt at t=0

• ESTIMATE y1 as

001

tdt

dytyy



Numerical Solution - 1 Notation Exact Numerical

Method (impossible to achieve) by Forward Steps

tntn

)( nn tyy

),( nnn ytff

Number Step n

Length Step Time t

),( ytfdtdy

Now Consider

yn+1

tn

yn

tn+1

tt



Numerical Solution - 2 The diagram at Left shows

that the relationship between yn, yn+1 and the CHORD slope

yn+1

tn

yn

tn+1

tt

slope chord 1

tyy nn

The problem with this formula is we canNOT calculate the CHORD slope exactly • We Know Only Δt & yn, but

NOT the NEXT Step yn+1

The AnalystChooses Δt

ChordSlope

Tangent Slope



Numerical Solution -3 However, we can

calculate the TANGENT slope at any point FROM the differential equation itself

The Basic Concept for all numerical methods for solving ODE’s is to use the TANGENT slope, available from the R.H.S. of the ODE, to approximate the chord slope

Recognize dy/dt as the Tangent Slope

),( nntt

n ytfdtdym

n

),( slopetangent ytf nnt

nn ytfdtdy

tyy

n

,1



Euler Method – 1st Order ODE Solve 1st Order

ODE with I.C. ReArranging

Use: [Chord Slope] [Tangent Slope at start of time step]

),( ytfdtdy

by )0( nnn ftyy 1

Then Start the “Forward March” with Initial Conditions

byt 00 0 nnt

nn ytfdtdy

tyy

n

,1

or1 nt

n ydtdyty

n



Example Euler Estimate Consider 1st

Order ODE with I.C.

Use The Euler Forward-Step Reln

See Next Slide for the 1st Nine Steps For Δt = 0.1

1 ydtdy

0)0( y

)1(1 nnn ytyy

ntn

nnn

dtdyty

ftyy

1

But from ODE

So In This Example:

1 nt

ydtdy

n



Euler Exmple Calcn tn yn fn= – yn+1 yn+1= yn+t fn

0 0 0.000 1.000 0.100

1 0.1 0.100 0.900 0.1902 0.2 0.190 0.810 0.271

3 0.3 0.271 0.729 0.344

4 0.4 0.344 0.656 0.410

5 0.5 0.410 0.590 0.469

6 0.6 0.469 0.531 0.522

7 0.7 0.522 0.478 0.570

8 0.8 0.570 0.430 0.613

9 0.9 0.613 0.387 0.651

1.01 tydtdy

Plot

Slope



Euler vs Analytical

0

0.2

0.4

0.6

0.8

0

0.25 0.5

0.75 1

1.25

t

Exact

Numerical

y

tey 1

The Analytical Solution



Analytical Soln Let u = −y+1 Then

001 tyydtdy

dudydydu

yu

10

1

Sub for y & dy in ODE

udtdu

Separate Variables

dtudu

Integrate Both Sides

dtudu

1

Recognize LHS as Natural Log

Ctu ln Raise “e” to the

power of both sidesCtu ee ln



Analytical Soln And

001 tyydtdy

Thus Soln u(t)tKeu

Sub u = 1−y

Now use IC

The Analytical Soln

ttCCt

u

Keeee

ue

ln

tKey 1

101 0

KKe

tey 11

tey 1



ODE Example: Euler Solution with

∆t = 0.25, y(t=0) = 37 The Solution Table

61.5ln2.4cos9.3 tydtdy

0 1 2 3 4 5 6 7 8 9 1022

24

26

28

30

32

34

36

38

t

y(t)

by E

uler

Euler Solution to dy/dt = 3.9cos(4.2y)-ln(5.1t+6)

n t y dy/dt dely yn+1

0 0 37.0000 -1.7457 -0.4364 36.56361 0.25 36.5636 1.4027 0.3507 36.91432 0.5 36.9143 -1.3492 -0.3373 36.57693 0.75 36.5769 1.2410 0.3103 36.88724 1 36.8872 -1.2264 -0.3066 36.58065 1.25 36.5806 1.0448 0.2612 36.84186 1.5 36.8418 -0.7108 -0.1777 36.66417 1.75 36.6641 1.1868 0.2967 36.96088 2 36.9608 -2.5004 -0.6251 36.33579 2.25 36.3357 -2.6357 -0.6589 35.6768

10 2.5 35.6768 -1.6265 -0.4066 35.270111 2.75 35.2701 0.0722 0.0181 35.288212 3 35.2882 -0.2436 -0.0609 35.227313 3.25 35.2273 0.4430 0.1107 35.338014 3.5 35.3380 -1.1420 -0.2855 35.052615 3.75 35.0526 -0.0139 -0.0035 35.049116 4 35.0491 -0.1072 -0.0268 35.022317 4.25 35.0223 -0.5255 -0.1314 34.890918 4.5 34.8909 -2.6041 -0.6510 34.239919 4.75 34.2399 -1.1497 -0.2874 33.952420 5 33.9524 -3.0108 -0.7527 33.199721 5.25 33.1997 -3.0006 -0.7502 32.449622 5.5 32.4496 -3.0151 -0.7538 31.695823 5.75 31.6958 -2.9862 -0.7466 30.949224 6 30.9492 -3.0384 -0.7596 30.189725 6.25 30.1897 -2.9328 -0.7332 29.456426 6.5 29.4564 -3.1419 -0.7855 28.671027 6.75 28.6710 -2.6916 -0.6729 27.998128 7 27.9981 -3.5484 -0.8871 27.111029 7.25 27.1110 -1.7458 -0.4365 26.674530 7.5 26.6745 -2.8722 -0.7180 25.956531 7.75 25.9565 -2.4562 -0.6141 25.342432 8 25.3424 -0.4717 -0.1179 25.224533 8.25 25.2245 -2.2562 -0.5641 24.660434 8.5 24.6604 -0.0369 -0.0092 24.651235 8.75 24.6512 -0.0977 -0.0244 24.626836 9 24.6268 -0.2699 -0.0675 24.559337 9.25 24.5593 -1.0481 -0.2620 24.297338 9.5 24.2973 -3.9863 -0.9966 23.300739 9.75 23.3007 -0.9318 -0.2329 23.067840 10 23.0678 -1.0551 -0.2638 22.8040



Compare Euler vs. ODE45Euler Solution ODE45 Solution

0 1 2 3 4 5 6 7 8 9 1022

24

26

28

30

32

34

36

38

t

y(t)

by E

uler


0 1 2 3 4 5 6 7 8 9 1034.5

35

35.5

36

36.5

37

37.5

T by ODE45

Y b

y O

DE

45

Euler is Much LESS accurate



Compare Again with ∆t = 0.025Euler Solution ODE45 Solution

0 1 2 3 4 5 6 7 8 9 1034.5

35

35.5

36

36.5

37

37.5

T by ODE45

Y b

y O

DE

45

Smaller ∆T greatly improves Result0 1 2 3 4 5 6 7 8 9 10

35.8

36

36.2

36.4

36.6

36.8

37

37.2

t

y(t)

by E

uler




MatLAB Code for Euler% Bruce Mayer, PE% ENGR25 * 04Jan11% file = Euler_ODE_Numerical_Example_1201.m%y0= 37;delt = 0.25;t= [0:delt:10]; n = length(t);yp(1) = y0; % vector/array indices MUST start at 1tp(1) = 0;for k = 1:(n-1) % fence-post adjustment to start at 0 dydt = 3.9*cos(4.2*yp(k))^2-log(5.1*tp(k)+6); dydtp(k) = dydt % keep track of tangent slope tp(k+1) = tp(k) + delt; dely = delt*dydt delyp(k) = dely yp(k+1) = yp(k) + dely;endplot(tp,yp, 'LineWidth', 3), grid, xlabel('t'),ylabel('y(t) by Euler'),... title('Euler Solution to dy/dt = 3.9cos(4.2y)-ln(5.1t+6)')



MatLAB Command Window forODE45

>> dydtfcn = @(tf,yf) 3.9*(cos(4.2*yf))^2-log(5.1*tf+6);>> [T,Y] = ode45(dydtfcn,[0 10],[37]);>> plot(T,Y, 'LineWidth', 3), grid, xlabel('T by ODE45'), ylabel('Y by ODE45')



Example Euler Approximation Use four steps of Δt = 0.1 with Euler’s

Method to approximate the solution to• With I.C.

SOLUTION: Make a table of values, keeping track

of the current values of t and y, the derivative at that point, and the projected next value.

10 ty



Example Euler Approximation Use I.C. to calculate the Initial Slope

Use this slope to Project to the NEW value of yn+1 = yn + Δy:

Then the NEW value for y:

210

11runrise1,0, 2

00

mdtdyyx

2.01.02 ytdtdytytm

2.12.01001 yyy



Example Euler Approximation Tabulating the remaining Calculations

The table then DEFINES y = f(t) Thus, for example, y(t=0.3) = 1.685

yhdtdy dtdy yy



WhiteBoard Work Problems From §9.4

• P32 Population Extinction

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&docid=mwI7LyqTNR2-oM&tbnid=smndac9x-ORZlM:&ved=0CAUQjRw&url=http://warnercnr.colostate.edu/~gwhite/pva/fslide45.htm&ei=V444U_OOGYaayQGA3YDgBA&bvm=bv.63808443,d.aWc&psig=AFQjCNFvu4KGyanUyDOagT_9apqOceJ92w&ust=1396301754935482



All Done for Today

CarlRunge

Carl David Tolmé Runge

Born: 1856 in Bremen, Germany

Died: 1927 in Göttingen, Germany



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

Appendix

–

srsrsr 22

a2 b2

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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Transcript of Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]