Assignment 5

Presented to: Prof. Taher G. Abu-El-Yazied

Prof. Niveen M. Badra

Presented by: Ahmed Hassan Ibrahim

1. (a) Use Euler's method with h = 0.25 to solve the following ODE between x = 0 and x = 1

(If the initial condition is y (0) = 1.0)

y ) x 1 ( xd

y d

𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖)ℎ

h 0.25

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑦𝑖+1

0 0 1 1 1.25

1 0.25 1.25 1.397542 1.599386

2 0.5 1.599386 1.897002 2.073636

3 0.75 2.073636 2.520022 2.703642

4 1 2.703642 3.288551 3.525779

𝑦4 = 2.703642

(b) Use the Heun's method with h = 0.25 to solve the previous ODE between x = 0 and x = 1

𝑦𝑖+1° = 𝑦

𝑖+ 𝑓(𝑥𝑖, 𝑦

𝑖)ℎ

𝑦𝑖+1 = 𝑦𝑖 +𝑓(𝑥𝑖 , 𝑦𝑖) + 𝑓(𝑥𝑖+1, 𝑦𝑖+1

°)

2ℎ

h 0.25

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑦𝑖+1° 𝑥𝑖+1 𝑓(𝑥𝑖+1, 𝑦𝑖+1

°) 𝑦𝑖+1

0 0 1 1 1.25 0.25 1.39754249 1.299693

1 0.25 1.299693 1.425051 1.655956 0.5 1.93025904 1.719107

2 0.5 1.719107 1.966721 2.210787 0.75 2.60202503 2.2902

3 0.75 2.2902 2.648346 2.952286 1 3.43644366 3.050798

4 1 3.050798 3.493307 3.924125 1.25 4.45711609 4.044601

𝑦4 = 3.050798

(c) Use the Midpoint method with h = 0.25 to solve the previous ODE between x = 0 and x = 1

𝑥𝑖+1/2 = 𝑥𝑖 +ℎ

2

𝑦𝑖+1/2 = 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖)ℎ

2

𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥+1/2𝑖 , 𝑦𝑖+1/2)ℎ

h 0.25

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑥𝑖+1/2 𝑦𝑖+1/2 𝑓(𝑥+1/2𝑖 , 𝑦𝑖+1/2) 𝑦𝑖+1

0 0 1 1 0.125 1.125 1.19324269 1.298311

1 0.25 1.298311 1.424293 0.375 1.47634729 1.6706942 1.715984

2 0.5 1.715984 1.964934 0.625 1.96160094 2.27592893 2.284966

3 0.75 2.284966 2.645318 0.875 2.61563126 3.03242125 3.043072

4 1 3.043072 3.48888 1.125 3.47918183 3.96367007 4.033989

𝑦4 = 3.050798

2. (a) Use Euler's method with h = 0.5 and h = 0.25 to solve: ( initial condition is y (0) = 1.0 )

xd

y d = x 2 y - 1.2 y

𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖)ℎ

h 0.5

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑦𝑖+1

0 0 1 -1.2 0.4

1 0.5 0.4 -0.38 0.21

2 1 0.21 -0.042 0.189

𝑦2 = 0.21

h 0.25

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑦𝑖+1

0 0 1 -1.2 0.7

1 0.25 0.7 -0.79625 0.500938

2 0.5 0.500938 -0.47589 0.381965

3 0.75 0.381965 -0.2435 0.321089

4 1 0.321089 -0.06422 0.305035

𝑦4 = 0.321089

(b) Use the Heun's method with h = 0.5 to solve the previous ODE.

𝑦𝑖+1° = 𝑦

𝑖+ 𝑓(𝑥𝑖, 𝑦

𝑖)ℎ

𝑦𝑖+1 = 𝑦𝑖 +𝑓(𝑥𝑖 , 𝑦𝑖) + 𝑓(𝑥𝑖+1, 𝑦𝑖+1

°)

2ℎ

h 0.25

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑦𝑖+1° 𝑥𝑖+1 𝑓(𝑥𝑖+1, 𝑦𝑖+1

°) 𝑦𝑖+1

0 0 1 -1.2 0.4 0.5 -0.38 0.605

1 0.5 0.605 -0.57475 0.317625 1 -0.063525 0.445431

2 1 0.445431 -0.08909 0.400888 1.5 0.42093253 0.528393

𝑦2 = 0.445431

(c) Use the Midpoint method with h = 0.5 and h = 0.25 to solve the previous ODE.

𝑥𝑖+1/2 = 𝑥𝑖 +ℎ

2

𝑦𝑖+1/2 = 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖)ℎ

2

𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥+1/2𝑖 , 𝑦𝑖+1/2)ℎ

h 0.25

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑥𝑖+1/2 𝑦𝑖+1/2 𝑓(𝑥+1/2𝑖 , 𝑦𝑖+1/2) 𝑦𝑖+1

0 0 1 -1.2 0.125 0.85 -1.00671875 0.74832

1 0.25 0.74832 -0.85121 0.375 0.64191852 -0.68003243 0.578312

2 0.5 0.578312 -0.5494 0.625 0.50963763 -0.41248796 0.47519

3 0.75 0.47519 -0.30293 0.875 0.4373235 -0.18996239 0.4277

𝑦4 = 0.47519

h 0.25

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑥𝑖+1/2 𝑦𝑖+1/2 𝑓(𝑥+1/2𝑖 , 𝑦𝑖+1/2) 𝑦𝑖+1

0 0 1 -1.2 0.125 0.7 -0.8290625 0.585469

1 0.5 0.585469 -0.5562 0.625 0.44641992 -0.36132112 0.404808

2 1 0.404808 -0.08096 1.125 0.38456778 0.02523726 0.417427

𝑦2 = 0.404808

3. Given that:

td

y d = 30 [sin (t) - y] + 3 cos (t)

With y (0) = 0. Use the implicit Euler to obtain a solution from t=0 to 4 using a step size of 0.4.

𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖)ℎ

h 0.5

i 𝑥𝑖 𝑦𝑖 𝑓(𝑥𝑖 , 𝑦𝑖) 𝑦𝑖+1

0 0 1 -27 -1.7

1 0.1 -1.7 56.98001 3.998001

2 0.2 3.998001 -111.04 -7.10598

3 0.3 -7.10598 224.9109 15.38511

4 0.4 15.38511 -447.108 -29.3256

𝑦4 = 15.38511

4. Given that:

td

y d = 999 y + 1999 z &

td

z d = - 1000 y – 2000 z

With y (0) = z (0) = 1. Obtain a solution between t=0 and 0.25 using a step size of 0.05 with the explicit

Euler method. Repeat your solution using implicit Euler method.

𝑦𝑖+1 = 𝑦𝑖 + 𝑓(𝑥𝑖 , 𝑦𝑖)ℎ

Implicit

𝑓𝑖𝑟𝑠𝑡 𝑒𝑞𝑎𝑢𝑡𝑖𝑜𝑛 ∶ y𝑖+1 = 𝑦𝑖 + (999 y𝑖+1 + 1999 z𝑖+1)(0.05)

𝑠𝑒𝑐𝑜𝑛𝑑 𝑒𝑞𝑎𝑢𝑡𝑖𝑜𝑛 ∶ 𝑧𝑖+1 = 𝑧𝑖 + (−1000 y𝑖+1 − 2000 z𝑖+1)(0.05)

𝑓𝑟𝑜𝑚 𝑓𝑖𝑟𝑠𝑡 𝑒𝑞𝑎𝑢𝑡𝑖𝑜𝑛: 𝑦𝑖 = −48.95 y𝑖+1 − 99.95 z𝑖+1

𝑓𝑟𝑜𝑚 𝑠𝑒𝑐𝑜𝑛𝑑 𝑒𝑞𝑎𝑢𝑡𝑖𝑜𝑛: 𝑧𝑖 = 50 y𝑖+1 + 101 z𝑖+1

y𝑖+1 = 1.88607 𝑦𝑖 + 1.86648 𝑧𝑖

z𝑖+1 = 0.9337 𝑦𝑖 − 0.9141 𝑧𝑖

h 0.5

i 𝑥𝑖 𝑦𝑖 𝑦𝑖+1 𝑧𝑖+1

0 1 1 3.75255 -1.8478

0.05 3.75255 -1.8478 3.62869 -1.81468

0.1 3.62869 -1.81468 3.456896 -1.72931

0.15 3.456896 -1.72931 3.292231 -1.64694

0.2 3.292231 -1.64694 3.135389 -1.56848

0.25 3.135389 -1.56848 2.986019 -1.49376

𝑦5 = −1.56848

5. The steady state heat balance for a rod is represented by:

x

d2

2

d

T - 0.1 T = 0.0

A B x

T (A) =200 T (B) =100

Where T is the temperature. Use the Finite difference method to solve the previous ODE for a bar

length of 10 m. Obtain the solution every 1.0 m.

𝑑2𝑦

𝑑𝑥2|

𝑖

=𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1

(∆𝑥)2

𝑑 𝑦

𝑑𝑥 |

𝑖

=𝑦𝑖+1 − 𝑦𝑖−1

2∆𝑥

𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1

(∆𝑥)2− 𝟎. 𝟏𝑻 = 𝟎

𝑇𝑖+1 − 2.1𝑇𝑖 + 𝑇𝑖−1 = 𝟎

i t1 t2 t3 t4 t5 t6 t7 t8 t9 =

1 -2.1 1 0 0 0 0 0 0 0 -200

2 1 -2.1 1 0 0 0 0 0 0 0

3 0 1 -2.1 1 0 0 0 0 0 0

4 0 0 1 -2.1 1 0 0 0 0 0

5 0 0 0 1 -2.1 1 0 0 0 0

6 0 0 0 0 1 -2.1 1 0 0 0

7 0 0 0 0 0 1 -2.1 1 0 0

8 0 0 0 0 0 0 1 -2.1 1 0

9 0 0 0 0 0 0 0 1 -2.1 -100

6. Use the finite difference method to obtain a solution for the following ODE between x=0 and x=10

with step size ∆x=2.0: ( The boundary conditions are y (0)=5 and y (10)=8 )

x

d2

2

d

y - 2

xd

y d - y + x = 0.0

𝑑2𝑦

𝑑𝑥2|

𝑖

=𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1

(∆𝑥)2

𝑑 𝑦

𝑑𝑥 |

𝑖

=𝑦𝑖+1 − 𝑦𝑖−1

2∆𝑥

𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1

(∆𝑥)2− 2

𝑦𝑖+1 − 𝑦𝑖−1

2∆𝑥− 𝑦𝑖 + 𝑥𝑖 = 0

𝑦𝑖+1 − 2𝑦𝑖 + 𝑦𝑖−1 − ∆𝑥(𝑦𝑖+1 − 𝑦𝑖−1) + (∆𝑥)2(−𝑦𝑖 + 𝑥𝑖) = 0

−1𝑦𝑖+1 − 6𝑦𝑖 + 3𝑦𝑖−1 + 4𝑥𝑖 = 0

i xi y1 y2 y3 y4 =

1 2 -6 -1 0 0 -23

2 4 3 -6 -1 0 -16

3 6 0 3 -6 -1 -24

4 8 0 0 3 -6 -24

(𝐀|𝐈) = (𝐈|𝐀−𝟏)

(

−6 −1 03 −6 −10 3 −6

00

−10 0 3 −6

) (

𝑦1

𝑦2𝑦3

𝑦4

) = (

−23−16−24−24

)

[𝐴][𝑌] = [𝐶] ⟹ [𝑌] = [𝐴]−1 [𝐶]

(

𝑦1

𝑦2𝑦3

𝑦4

) = (

−0.15469611 0.023941 −0.00368

−0.07182321 −0.14365 0.022099

−0.03314916 −0.0663 −0.14365

0.000614

−0.00368

0.023941−0.01657458 −0.03315 −0.07182 −0.1547

) (

−23−16−24−24

)

(

𝑦1

𝑦2𝑦3

𝑦4

) = (

3.248539

3.5082784.696247

6.348095

)