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NUMERICAL ANALYSIS

NUMERICAL ANALYSIS. Course for 2nd year Architecture Engineering. By:. Dr Rania Bahgat Mohamed Amer. Department of Engineering Mathematics and Physics, Faculty of Engineering Zagazig University 2009-2010. CONTENTS OF THE COURSE:. Numerical solution of linear equations.

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NUMERICAL ANALYSIS

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  1. NUMERICAL ANALYSIS Course for 2nd year Architecture Engineering

  2. By: Dr\ Rania Bahgat Mohamed Amer Department of Engineering Mathematics and Physics, Faculty of Engineering Zagazig University 2009-2010

  3. CONTENTS OF THE COURSE: Numerical solution of linear equations. Numerical solution of non linear equations. Numerical integrations. Curve fitting. Numerical solution of ordinary differential equations. Numerical solution of partial differential equations.

  4. Next The Main Menu اPrevious Numerical solution of ordinary differential equations 1- Introduction 2- Taylor Method 3- Euler Method 4- Modified Euler Method 5- Rung-Kutta Method

  5. Next The Main Menu اPrevious 1- Introduction Numerical ordinary differential equations is the part of numerical analysis which studies the numerical solution of ordinary differential equations (ODEs). It is very important forindustrial and practical engineering applications ,Mechanics, Chemistry and Physics. Boundary value problem It is an ordinary differential equation which satisfies the boundary conditions. A boundary value problem has conditions specified at the extremes of the independent variable Initial value problem It is an ordinary differential equation together with specified value, called the initial condition, of the unknown function at a given point in the domain of the solution.Initial value problem has all of the conditions specified at the same value of the independent variable in the equation and that value is at the lower boundary of the domain .

  6. Next The Main Menu اPrevious 2- Taylor Method It is a numerical methods for solving ordinary differential equations (O.D.Es) with a given conditions. y\ = f(x, y), y(x0) = y0. Taylor series expansion of the function f(x) around a point of expansion x has the following form: Assume h = x – x0 Hint: 1- When the step h becomes small, the numerical solution becomes nearly the same as the exact solution. 2- y = f(x). 3- computing approximate values of the solution f(x) at the points: x1 = x0 +h, x2 = x0 +2h, x3 = x0 + 3h,…………….

  7. Next The Main Menu اPrevious Example (1) Solve the following boundary value problem y\ = - y at x = 0.2, 0.4 given that y(0) = 1 Solution Taylor expansion given by: Consider the point of expansion zero Evaluate the derivatives at the point of expansion zero Then

  8. Next The Main Menu اPrevious 3- Euler Method It is a first-order numerical procedure for solving ordinary differential equations (O.D.Es) with a given conditions . y\ = f(x, y), y(x0) = y0. The general Euler formula for solving boundary value problem is: Assume h = xn+1 – xn Hint: 1- When the step h becomes small, the numerical solution becomes nearly the same as the exact solution. 2- y = f(x). 3- computing approximate values of the solution f(x) at the points: x1 = x0 +h, x2 = x0 +2h, x3 = x0 + 3h,…………….

  9. Next The Main Menu اPrevious Example (1) Solve the following boundary value problem y\= x2 + y at x = 0.02, 0.04, 0.06 and 0.08 given that y(0) = 1 . Solution For n = 1 General Euler formula given by: Then For n = 0 Then

  10. Next The Main Menu اPrevious For n = 2 Then For n = 3 Then For n = 4 Then

  11. Next The Main Menu اPrevious We can put the results in a table:

  12. Next The Main Menu اPrevious 4- Modified Euler Method This method tries to improve the error of Euler method. The general modified Euler formula for solving boundary value problem is: and Euler method This method is an iterative method, starts by getting y01 using Eulermethod. Then we calculate f (xn+1,y0n+1 ). Then determine y11. After that we use y11 to get the new value y21 and so on. The solution is terminated after a certain number of steps or certain accuracy required.

  13. Next The Main Menu اPrevious Example (1) Find y(0.1) and y(0.2), if y\ = y – x , y(0)= 2, take h = 0.1, three decimal places are required Solution To get y1 = y(0.1) Then y(0.1) = 2.205 To get y2 = y(0.2) Then y(0.2) = 2.241

  14. Next The Main Menu اPrevious 5- Rung-Kutta Method The accuracy of this method is equivalent to five terms of Talyor’s series. The general Rung- kutta formula for solving boundary value problem is: where

  15. Next The Main Menu اPrevious Example (1) Find y(0.1) and y(0.2), if y\ = y – x , y(0)= 2, take h = 0.1, four decimal places are required Solution To get y1 = y(0.1)

  16. Next The Main Menu اPrevious Then y(0.1) = 2.2052 To get y2 = y(0.2) Then y(0.2) = 2.4215

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