Question

Question 4 A mathematical model has been described by an engineer into the following differential equation: dy 0.5x0 dx y(0) 2.5 Demonstrate an Euler method simulation of y versus x with a tabular algorithm using a. 0.5 and 0.0 3.0. x Demonstrate a 4th-order Runge Kutta method simulation of y versus x with a tabular b. algorithm using дх-0.5 and 0.0 XS 3.0. What can you say about y(x) and the methods used? c.

Answer 1

a) Euler’s Method to solve ODE

y – y’ + 0.5x = 0

i.e    y’ = y + 0.5x ;             at x_o = 0, y₀ = 2.5

thus, f(x,y) = y + 0.5x

for h = 0.5

at x_o = 0, y₀ = 2.5

Now, f(x_o , y₀ )= 2.5 + 0.5(0) = 2.5

At x₁ = 0.5,          y₁ = y₀ +h[f(x_o , y₀ )] = 2.5 + 0.5(2.5) = 3.75

Now, f(x₁ , y₁ ) = 3.75 + 0.5(0.5) = 4

At x₂ = 1,              y₂ = y₁ +h[f(x₁ , y₁ )] = 3.75 + 0.5(4) = 5.75

We continue iterating in excel and tabulate the result below

x	f(x,y)	y
0	2.5	2.5
0.5	4	3.75
1	6.25	5.75
1.5	9.625	8.875
2	14.6875	13.6875
2.5	22.2813	21.0313
3	33.6719	32.1719

35 30 25 20 15 10 0.5 1.5 2.5 3.5

b) Runge Kutte of t order

y – y’ + 0.5x = 0

i.e    y’ = y + 0.5x ;            

thus, f(x,y) = y + 0.5x

Given : x_o = 0, y₀ = 2.5     and h =0.5

Iteration 1: at x = x₀ + h = 0+0.5 = 0.5

K₁ =hf(x₀ , y₀) = 0.5[2.5 + 0.5 (0)] = 1.25

K₂ =hf(x₀ +h/2, y₀ + k₁ /2) = 0.5 *f(0.25, 3.125) = 1.625

K₃ =hf(x₀ +h/2, y₀ + k₂ /2) =0.5 *f(0.25, 3.3125) = 1.71875

K₄ =hf(x₀ +h, y₀ + k₃ ) = 0.5 *f(0.5, 4.21875) = 2.2344

y₁ = y₀ + 1/ 6 (k₁ + 2k₂ + 2k₃ + k₄) = 2.5 + 1/6 (1.25 + 2x1.625 + 2x1.7188 + 2.2344) = 4.1953

We carry on with further iteration in excel and the results are as shown below:

x	y	k1	x0 + h/2	y0+k1/2	k2	yo+k2/2	k3	x0+h	y0+k3	k4
0	2.5000	1.2500	0.2500	3.1250	1.6250	3.3125	1.7188	0.5000	4.2188	2.2344
0.5	4.1953	2.2227	0.7500	5.3066	2.8408	5.6157	2.9954	1.0000	7.1907	3.8453
1	7.1520	3.8260	1.2500	9.0650	4.8450	9.5746	5.0998	1.5000	12.2518	6.5009
1.5	12.1881	6.4691	1.7500	15.4227	8.1488	16.2625	8.5688	2.0000	20.7569	10.8784
2	20.6519	10.8260	2.2500	26.0649	13.5949	27.4494	14.2872	2.5000	34.9391	18.0946
2.5	34.7660	18.0080	2.7500	43.7701	22.5725	46.0523	23.7137	3.0000	58.4797	29.9898
3	58.1944

70.0000 60.0000 50.0000 40.0000 30.0000 20.0000 10.0000 0.0000 0.5 2.5 3.5

c) The curve y(x) is a smooth second degree curve in both the solutions

Runge Kutta method gives a much smoother approximation of the solution as compared to the Euler's method.