FACULTY OF CIVIL ENGINEERING AND ENVIRONMENTAL

SEMESTER 2 2014/ 2015

MATHEMATICS ENGINEERING IV

DR. JOEWONO PRASETIJO

ELYA SYAHERA BINTI MOHD SALLEH

AF 130067

940111-01-5564

h = 0.2

Solve the value boundary problem y” – xy’ + 2y = 10x, 1< x < 2 , with conditions

y (1) = 1.5 y (2) = 15 . Use h = 0.2

Solution :

x0=1 x1 = 1.2 x2 = 1.4 x3 = 1.6 x4 = 1.8 x5 = 2.0

Find y0 , Given that y(1 ) = 1.5, hence y0 = 1.5

Find y”, Given that y(2 ) = 15, hence y5 = 15

yi” – xiyi’ + 2yi - 10xi = 0

y i+1−2 y 'i+ y i−1

2 h−x ( y i+1− y i−1

2h )+2 y i−10 x i= 0

x h2 yi+1 -2 yi + y i-1 - h2

xi (y i+1 – yi-1 ) + 2 h yi – 10 xi = 0

Let h = 0.2 , yi+1 -2 yi + yi-1 – 0.1 xi (yi+1 – yi-1 ) +0.08 yi – 0.4 xi = 0

Rearrange, ( 1 + 0.1 xi) yi-1 – 1.92 yi + ( 1- 0.1xi )yi+1 = 0.4xi

Substitute i = 1, 2, …n-1

For i = 1, ( 1 + 0.1 x1) y0 – 1.92 y1 + ( 1- 0.1x1 )y2= 0.4x1

Let xi = 1.2, y0 = 1.5, 1.12 (1.5) – 1.92 y1 + 0.88y2 = 0.48

-1.92 y1 + 0.88 y2 = -1.20

For i = 2, ( 1 + 0.1 x2) y1 – 1.92 y2 + ( 1- 0.1x2 )y3= 0.4x2

Let xi = 1.4 ( 1 + 0.1 (1.4))y1 – 1.92 y2 + (1 – 0.1(1.4))y3 = 0.4(1.4)

-1.14y1 – 1.92 y2 + 0.86 y3 = 0.56

For i = 3, ( 1 + 0.1 x3) y2– 1.92 y3+ ( 1- 0.1x3 )y4= 0.4x3

Let xi = 1.6 ( 1 + 0.1 (1.6))y2 – 1.92 y3 + (1 – 0.1(1.6))y4 = 0.4(1.6)

1.16 y2 – 1.92 y3 + 0.84 y4 = 0.64

For i = 4, ( 1 + 0.1 x4 ) y3 – 1.92 y4+ ( 1- 0.1x4 )y5= 0.4x4

Let xi = 1.8 ( 1 + 0.1 (1.8))y3 – 1.92 y4 + (1 – 0.1(1.8))y5 = 0.4(1.8)

1.18y3 – 1.92 y4 + 0.82 y5 = 0.72

y5 = 15 1.18y3 – 1.92 y4 + 0.82 (15)= - 11.58

1.18y3 – 1.92 y4 = - 11.58

In matrix – vector form

(−1.92 0.88 01.14 −1.92 0.86

0 1.16 −1.920 0 1.18

00

0.84−1.92

)(y1

y2

y3

y4

)=(−1.200.560.64

−11.58)