BMayer@ChabotCollege.edu MTH16_Lec-14_sec_9-4_ODE_SlopeFields_Euler.pptx 1 Bruce Mayer, PE Chabot...

BMayer@ChabotCollege.edu • MTH16_Lec-14_sec_9-4_ODE_SlopeFields_Euler.pptx 1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

§9.4 ODEAnalytic

Review §

Any QUESTIONS About• §9.3 Differential Equation Applications

Any QUESTIONS About HomeWork• §9.3 → HW-15

§9.4 Learning Goals

Analyze solutions ofdifferential 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

dxxdy 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

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:

ydxxdyxdx

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

-2 -1 1 2

Slope Fields

Of course this Variable Separable ODE can be easily solved analytically

dy dxxdyy

dxxdyy

Cxy 22

Cyx 22

-2 -1 1 2

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

dy dxxdy

Cxxy 2

C 2 24

1 2 xxySoln

Slope Field Identification

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

22 yxdx

1 yydx

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.

SOLUTION: Calculate some

Slope Values from peDm

1100,0 0 emdp

0111,1 1 emdp

068.015.05.0,2 2 emdp

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

$k/unit p

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

Removing ABS Bars

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

CeCeDp eeDeeCeDpp

11ln 1ln

pp eeC eADDeeD 110with 1

572.05.02

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

Numerical Solution - 1

Notation Exact Numerical Method (impossible to achieve) by Forward Steps

)( nn tyy

),( nnn ytff

Number Step n

Length Step Time t

),( ytfdt

Now Consider

Numerical Solution - 2 The diagram at Left shows

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

slope chord 1

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 ytfdt

),( slopetangent ytf nnt

nn ytfdt

Euler Method – 1st Order ODE

Solve 1st Order ODE with I.C.

ReArranging

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

),( ytfdt

by )0( nnn ftyy 1

Then Start the “Forward March” with Initial Conditions

byt 00 0 nnt

nn ytfdt

or1 nt

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

0)0( y

)1(1 nnn ytyy

But from ODE

So In This Example:

Euler Exmple Calc

n tn yn fn= – yn+1 yn+1= yn+Dt fn

0 0 0.000 1.000 0.100

1 0.1 0.100 0.900 0.190

2 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 tydt

Euler vs Analytical

Numerical

The Analytical Solution

Analytical Soln

Let u = −y+1 Then

001 tyydt

Sub for y & dy in ODE

Separate Variables

Integrate Both Sides

Recognize LHS as Natural Log

Ctu ln

Raise “e” to the power of both sides

Ctu ee ln

Analytical Soln

001 tyydt

Thus Soln u(t)tKeu

Sub u = 1−y

Now use IC

The Analytical Soln

tKey 1

tey 11

ODE Example: Euler Solution with

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

61.5ln2.4cos9.3 tydt

0 1 2 3 4 5 6 7 8 9 1022

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

0 1 2 3 4 5 6 7 8 9 1034.5

T by ODE45

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

T by ODE45

Smaller ∆T greatly improves Result

0 1 2 3 4 5 6 7 8 9 1035.8

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.

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:

rise1,0, 2

2.01.02 ytdt

dytytm

2.12.01001 yyy

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

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

BMayer@ChabotCollege.edu

Chabot Mathematics

Appendix

srsrsr 22

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