1. Numerical Solutions of Ordinary Differential Equations
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2. Ordinary Differential Equations
Equations which are composed of an unknown function and its derivatives are called differential equations.
Differential equations play a fundamental role in engineering because many physical phenomena are best formulated mathematically in terms of their rate of change.
v - dependent variable
t - independent variable
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3. Types of Differential Equations
When a function involves one independent variable, the equation is called an ordinary differential equation (ODE).
A partial differential equation (PDE) involves two or more independent variables.
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4. Model Ordinary Differential Equations
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5. Why Numerical solutions of ODE ?
Many differential equations cannot be solved analytically, in which case we have to satisfy ourselves with an approximation to the solution.
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6. Initial value problem
Initial value problem associated with first order differential equations is
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7. Methods of solving IVP Taylor series Picard successive approximation
Euler
Modified Euler
Runge-Kutta (RK)
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8. Taylor Series Method
The Taylor series method is a straight forward adaptation of classic calculus to develop the solution as an infinite series.
The method is not strictly a numerical method but it is used in conjunction with numerical schemes.
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9. Taylor Series Method
The Taylor series expansion of y(x) about x = x0 is
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10. Problem#1
The differential equation y'(x) = x + y satisfying y(0)=1. Determine the Taylor series solution. Also compare with the exact solution. {Analytical solution: y(x)=2ex - x -1(Cauchy linear differential equation).}
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11. Solution First, we find derivatives of y' = x + y, 4/17/2017
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12. Solution The Taylor series expansion of y(x) about x = x0 is 4/17/2017
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13. Comparison of solutions
The exact solution is Therefore, Taylor series solution and exact solution are equal.
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14. Picard’s Method The initial value Problem
is equivalent to the following integral equation
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15. Picard’s Method The Picard’s nth approximate solution is 4/17/2017
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16. Problem#1
Using Picard’s process of successive approximation, obtain a solution up to the fifth approximation of the equation
such that y = 1 when x = 0. Check your answer by finding the exact solution.
Ans.
Exact solution y(x) = 2ex – (x+1).
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17. Application Problem
For the free-falling bungee jumper with linear drag, compute the velocity at 10 seconds of a free-falling parachutist of mass 80kg and drag coefficient 10 kg/s at 10 seconds, with initial velocity 20 m/s, using (i) Taylor series method (ii) Picard’s method.
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18. Solution
Let v(t) be the velocity and ‘a’ be the acceleration of the parachutist at any time t.
Then the total force acting on the parachutist is equal to the sum of the
downward force FD due to gravity and
upward force FU due to air resistance.
i.e. F = FD + FU
FD = mg
FU α v (linear drag) FU = - Cd v(t).
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19. Solution
By Newton second law of motion F = ma. ma = mg – Cd v Therefore,
Given that m = 80kg, Cd = 10 kg/s,
and initial velocity v(0)=20 m/s.
Now solve the above IVP using Taylor and Picard’s method.
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20. Euler’s Method
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21. Interpretation of Euler Methody2 y1 y0
x x x x
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22. Interpretation of Euler Method
Slope=f(x0,y0) y1
y1=y0+hf(x0,y0)
hf(x0,y0) y0
x x x x h
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23. Interpretation of Euler Methody2
y2=y1+hf(x1,y1)
Slope=f(x1,y1)
hf(x1,y1)
Slope=f(x0,y0) y1
y1=y0+hf(x0,y0)
hf(x0,y0) y0
x x x x h h
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24. Interpretation of Euler Method
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25. Problem#1
Using Euler’s method, find an approximate value of y corresponding to x=1, given that yʹ = x + y such that y=1 when x=0. Ans: Taking h = 0.1, then y(0.1) = 1.1, y(0.2) = 1.22, y(0.3) = 1.36 y(0.4) = 1.53, y(0.5) = 1.72, y(0.6) = 1.94 y(0.7) = 2.19, y(0.8) = 2.48, y(0.9) = 2.81 y(1.0) = 3.18.
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26. Problem#2
A cup of a coffee originally has temperature of 68oC. The temperature of the ambient is 21oC and the thermal constant is 0.017oC/min. Determine the temperature of the coffee from t = 0 to 10 minutes insteps of 2 minutes. Sol. The differential equation is Tʹ(t) = -k(T - Ta) = (T - 21) T(0) = 68
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27. Improvements of Euler’s method
A fundamental source of error in Euler’s method is that the derivative at the beginning of the interval is assumed to apply across the entire interval.
Two simple modifications are available to circumvent this shortcoming:
Heun’s (Modified Euler) Method
Midpoint (Improved Polygon) Method
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28. Modified Euler’s Method
To improve the estimate of the slope, determine two derivatives for the interval:
(1) At the initial point
(2) At the end point
The two derivatives are then averaged to obtain an improved estimate of the slope for the entire interval.
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29. Euler Modified method
To improve the estimate of the slope, determine two derivatives for the interval:
At the initial point
At the end point
The two derivatives are then averaged to obtain an improved estimate of the slope for the entire interval.

30. Heun’s method (improved)
Original Huen’s:
Note that the corrector can be iterated to improve the accuracy of yi+1 .

31. Problem#1
Using Modified Euler’s method, find an approximate value of y when x= 0.3, given that yʹ = x + y and y=1 when x=0.
Sol. Compare the IVP with the general IVP
yʹ = f(x, y), y(x0) = x0, we have
f(x, y) = x + y, x0 = 0 and y0 = 1.
Taking the step size h = 0.1, then x0 = 0, x1 = 0.1, x2 = 0.2, x3 = 0.3. Now we find the corresponding value of y, using modified Euler’s method.

32. Solution Step-1. To find y1 = y(x1) = y(0.1)
Predictor formula (Euler’ formula)
y1(0) = y0 + h f(x0, y0)
= f(0, 1) = 1+0.1(0+1) =1.1
Corrector formula
y1(k) = y0 + (h/2) [f(x0, y0) + f(x1, y1(k-1))]
y1(1) = y0 + (h/2) [f(x0, y0) + f(x1, y1(0))]
= 1 + (0.1/2) [f(0,1) + f(0.1, 1.1)]
= [ ] =1.11

33. Solution y1(2) = y0 + (h/2) [f(x0, y0) + f(x1, y1(1))]
= 1 + (0.1/2) [f(0,1) + f(0.1, 1.11)]
= [ ] =
y1(3) = y0 + (h/2) [f(x0, y0) + f(x1, y1(2))]
= 1 + (0.1/2) [f(0,1) + f(0.1, )]
= [ ]
Therefore, y1 =

34. Solution Step-2. To find y2 = y(x2) = y(0.2)
Predictor formula (Euler’ formula)
y2(0) = y1 + h f(x1, y1)
= f(0.1, )
= ( ) =
Corrector formula
y2(k) = y1 + (h/2) [f(x1, y1) + f(x2, y2(k-1))]
y2(1) = y1 + (h/2) [f(x1, y1) + f(x2, y2(0))]
= (0.1/2) [f(0.1,1.1105) + f(0.2, )]
= [ ] =

35. Solution y2(2) = y1 + (h/2) [f(x1, y1) + f(x2, y2(1))]
= (0.1/2) [f(0.1,1.1105) + f(0.2, )]
= [ ] =
y2(3) = y1 + (h/2) [f(x1, y1) + f(x2, y2(2))]
= (0.1/2) [f(0.1,1.1105) + f(0.2, )]
= [ ]
Therefore, y2 = y(0.2)=
Similarly, y3 = y(0.3) =

36. Application Problem
A storage tank (Fig.) contains a liquid at depth y where y = 0 when the tank is half full. Liquid is withdrawn at a constant flow rate Q=500m3/h to meet demands. The contents are resupplied at a sinusoidal rate 3Q sin2(t). The surface area of the tank is 1200 m2. Determine the depth of the tank from t= 0 to t=6 in steps of 2 hours, using modified Euler’s method.
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37. Problem#1 Q =500 m3/h 3Q sin2(t) 4/17/2017
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38. Solution
Let y(t) be the depth of the tank at any time t and A be the surface area of the tank. The depth of the tank is changed when volume of the liquid in the tank changed. Therefore,
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39. Solution
Since A is constant, it implies that Given that Q=500m3/h, A =1200 m2 and y(0) = 0. Substituting these value, we have
Now solving the above initial value problem, using modified Euler’s method.
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40. Runge - Kutta Method
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41. Martin Wilhelm Kutta (1867-1944)
Carl Runge ( )
Martin Wilhelm Kutta ( )
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42. Introduction
Runge-Kutta methods are very popular because of their good efficiency; and are used in most computer programs for differential equations.
These methods agree with Taylor’s series solution up to the term in hr, where r differs method to method and is called the order of the method.
Euler’s and modified Euler’s are called the first and second order Runge-Kutta methods.
The fourth order R-K method is most commonly used and is often referred to as R-K method only.
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43. Working Rule
To find increment ‘k’ of y corresponding to an increment h of x by R-K method from the initial value problem
y´ = f(x, y) with y(x0) = y0 is as follows
Calculate
k1 = h f(x0, y0)
k2 = h f(x0+0.5h, y0+0.5k1)
k3 = h f(x0+0.5h, y0+0.5k2)
k4 = h f(x0+h, y0+k3)
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44. Working Rule Finally compute k = 1/6 (k1 + 2k2 + 2k3 + k4)
is the weighted mean of k1, k2, k3 and k4.
Therefore the required approximate value is
y1 = y(x1) = y(x0 + h) = y0 + k.
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45. Graphical Representationk2 k4 k3 k1 xi
xi + h/2
xi + h
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46. Problem#1
Using Runge-Kutta method of fourth order, solve yʹ = (y2 – x2)/(y2 + x2) with y(0)=1 at x = 0.2, 0.4.
Sol : - Compare the given problem with the general first order initial value problem yʹ=f(x, y) satisfying y(x0)=y0, we have
f(x, y) = (y2 – x2)/(y2 + x2)
x0 = 0, y0 =1.
Here we have to find value of y at x=0.2, 0.4.
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47. Solution
Taking step size h = 0.2. Step(1) To find y(0.2) k1 = h f(x0, y0) = 0.2 f(0, 1) = 0.2 (12-02)/(12+02) = 0.2 k2 = h f(x0+0.5h, y0+0.5k1) = 0.2 f(0+0.5(0.2), (0.2)) = 0.2 f(0.1, 1.1) =
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48. Solution
k3 = h f(x0+0.5h, y0+0.5k2) = 0.2 f(0+0.5(0.2), 1+0.5( )) = 0.2 f(0.1, ) = k4 = h f(x0+h, y0+k3) = 0.2 f(0+0.2, ) = 0.2 f(0.2, ) =
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49. Solution
k = 1/6 (k1 + 2k2 + 2k3 + k4) = 1/6 (0.2 +2( )+2(0.1967) ) = Hence y1 = y0 + k y(0.2) = = Step(2) To find y(0.4) k1 = h f(x1, y1) = 0.2 f(0.2, ) =
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50. solution
k2 = h f(x1+0.5h, y1+0.5k1) = 0.2 f( (0.2), (0.1891)) = 0.2 f(0.3, ) = k3 = h f(x1+0.5h, y1+0.5k2) = 0.2 f( (0.2), (0.1795)) = 0.2 f(0.3, ) =
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51. Solution
k4 = h f(x1+h, y1+k3) = 0.2 f( , ) = 0.2 f(0.4, ) = k = 1/6 (k1 + 2k2 + 2k3 + k4) = 1/6 ( (0.1795)+2(0.1793) ) = Hence y2 = y1 + k y(0.4) = =
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52. Problem#2
Apply Runge-Kutta method to find an approximate value of y for x=0.2 in steps of 0.1, if yʹ = x + y2 given that y=1 where
x = 0.
Sol : - Compare the given problem with the general first order initial value problem yʹ=f(x, y) satisfying y(x0)=y0, we have
f(x, y) = x + y2 , x0 = 0, y0 =1.
Here we have to find value of y at x = 0.2.
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53. Solution
Taking step size h = 0.1. Step(1) To find y(0.1) k1 = h f(x0, y0) = 0.1 f(0, 1) = 0.1 (0 + 12) = 0.1 k2 = h f(x0+0.5h, y0+0.5k1) = 0.1 f(0+0.5(0.1), (0.1)) = 0.2 f(0.05, 1.05) =
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54. Solution
k3 = h f(x0+0.5h, y0+0.5k2) = 0.1 f(0+0.5(0.1), 1+0.5(0.1152)) = 0.1 f(0.05, ) = k4 = h f(x0+h, y0+k3) = 0.1 f(0+0.1, ) = 0.1 f(0.1, ) =
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55. Solution
k = 1/6 (k1 + 2k2 + 2k3 + k4) = 1/6 (0.1 +2(0.1152)+2( ) ) = ≈ Hence y1 = y0 + k y(0.1) = = Step(2) To find y(0.2) k1 = h f(x1, y1) = 0.1 f(0.1, ) =
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56. solution
k2 = h f(x1+0.5h, y1+0.5k1) = 0.1 f( (0.1), (0.1347)) = 0.2 f(0.15, ) = k3 = h f(x1+0.5h, y1+0.5k2) = 0.1 f( (0.1), (0.1551)) = 0.1 f(0.15, 1.194) =
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57. Solution
k4 = h f(x1+h, y1+k3) = 0.1 f( , ) = 0.1 f(0.2, ) = k = 1/6 (k1 + 2k2 + 2k3 + k4) = 1/6 ( (0.1551)+2(0.1576) ) = Hence y2 = y1 + k y(0.2) = =
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