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Lec._9BUE

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    1

    ScheduleWeek Date Topic Classification of Topic

    1 9 Feb. 2010 Introduction toNumerical Methods

    and Type of Errors

    Measuring errors, Binary representation,

    Propagation of errors and Taylor series

    2 14 Feb. 2010 Nonlinear Equations Bisection Method

    3 21 Feb. 2010 Newton-Raphson Method

    4 28 Feb. 2010 Interpolation Lagrange Interpolation

    5 7 March 2010 Newton's Divided Difference Method

    6 14 March 2010 Differentiation Newton's Forward and BackwardDivided Difference

    7 21 March 2010 Regression Least squares

    8 28 March 2010 Systems of LinearEquations

    Gaussian Jordan

    9 11 April 2010 Gaussian Seidel

    10 18 April 2010 Integration Composite Trapezoidal and SimpsonRules

    11 25 April 2010 Ordinary DifferentialEquations

    Euler's Method

    12 2 May 2010

    Runge-Kutta 2nd and4th order Method

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    2

    Exact solution to Separable andLinear differential equations

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    3

    Ordinary differential equations of order one and degree one

    1. Separable equations

    Example 1

    Solve the equation2

    dy x

    dx y given that (0) 1y

    solution

    By separating the variables of the equation we get

    2y dy xdx

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    4

    By integration

    2y dy x dx then

    3 2

    3 2y x C The general solution of the equation

    Now (0) 1y y=1 at x=0

    then1 0

    3 2C

    1

    3C

    the solution of the equation is

    3 2 1

    3 2 3

    y x

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    5

    2. Linear equations

    Linear equations are on the form

    ( ) ( )

    dy

    p x y q xdx

    Example

    212

    dy y xdx x

    2

    1( )

    2( )

    p x

    x

    q x x

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    6

    How to solve Linear equations

    Linear equations are solved by using the following method

    Step 1: Calculate the following integrating factor

    ( )p x dx

    e

    Step 2: The solution is then given by

    ( )y q x dx

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    Example 2

    Solve the equation such that y(0)=244 ydx

    dy

    solution

    This equation could be written on the form

    44 ydx

    dy

    4)(

    4)(

    xq

    xp

    Thenxdxdxxp eee 4

    4)(

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    The solution will be such that

    ( )y q x dx cedxeye xxx

    444 4

    Let x=0 , y=2 c=3

    Thenxey 431

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    Euler Method

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    Eulers Method

    Step size, h

    x

    y

    x0,y0

    True value

    y1, Predicted

    value

    00,, yyyxfdx

    dy

    Slope

    Run

    Rise

    01

    01

    xx

    yy

    00 ,yxf

    010001 , xxyxfyy

    hyxfy 000 ,Figure 1 Graphical interpretation of the first step of Eulers method

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    Eulers Method

    Step size

    h

    True Value

    yi+1, Predicted value

    yi

    x

    y

    xi xi+1

    Figure 2. General graphical interpretation of Eulers method

    hyxfyy iiii ,1

    ii xxh 1

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    How to write Ordinary Differential

    Equation

    Example

    50,3.12 yeydx

    dy x

    is rewritten as 50,23.1 yye

    dx

    dy x

    In this case

    yeyxf x 23.1,

    How does one write a first order differential equation in the form of

    yxfdx

    dy,

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    ExampleA ball at 1200K is allowed to cool down in air at an ambient temperature of

    300K. Assuming heat is lost only due to radiation, the differential equation for

    the temperature of the ball is given by

    Kdtd 12000,1081102067.2 8412

    Find the temperature at 480t seconds using Eulersmethod. Assume a step size of

    240h seconds.

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    Solution

    K

    f

    htf

    htf iiii

    09.106

    2405579.4120024010811200102067.21200

    2401200,01200

    ,

    ,

    8412

    0001

    1

    Step 1:

    1is the approximate temperature at 240240001 httt

    K09.106240 1

    8412 1081102067.2 dt

    d

    8412

    1081102067.2,

    tf

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    Solution ContFor 09.106,240,1 11 ti

    K

    f

    htf

    32.110

    240017595.009.106

    240108109.106102067.209.10624009.106,24009.106

    ,

    8412

    1112

    Step 2:

    2 is the approximate temperatureat 48024024012 httt

    K32.110480 2

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    Comparison of Exact and

    Numerical Solutions

    Figure 3. Comparing exact and Eulers method

    0

    200

    400

    600

    800

    1000

    1200

    1400

    0 100 200 300 400 500

    Time, t(sec)

    Temperature,

    h=240

    Exact Solution

    (K

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    Step, h (480) Et |t|%

    480

    240

    120

    60

    30

    987.81

    110.32

    546.77

    614.97

    632.77

    1635.4

    537.26

    100.80

    32.607

    14.806

    252.54

    82.964

    15.566

    5.0352

    2.2864

    18

    Effect of step sizeTable 1. Temperature at 480 seconds as a function of step size, h

    K57.647)480( (exact)

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    Comparison with exact results

    -1500

    -1000

    -500

    0

    500

    1000

    1500

    0 100 200 300 400 500

    Time , t (sec)Temperature,

    Exact solution

    h=120h=240

    h=480

    (K

    Figure 4. Comparison of Eulers method with exact solution for different step sizes

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    Effects of step size on Eulers

    Method

    -1200

    -800

    -400

    0

    400

    800

    0 100 200 300 400 500

    Step size, h (s)Temperature,

    (K

    Figure 5. Effect of step size in Eulers method.

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