Question

Consider the following initial value problem:

1. Use Euler's explicit scheme to solve the above initial value problem with time step h= 0.5. Express all the computed results with a precision of three decimal places.

2. The analytical or exact solution is compute the absolute error at each t_ivalue. Express all the computed results with a precision of three decimal places.

3. Write a matlab function that solves the above (IVP) using (RK2.M) for arbitrary time-step h.

y(t) ly(0) 3 1.2y7e0.3 te [0, 1.5 43 -1.2t 70 -0.3t -e y = 9 -e 9

Answer 1

-. 2 teo,1.5] 3 Euler method (1) 0.3t fa.9)12y7 h-0.5 at to-0, yo-3 J@>3 Stepo 0. 3x 0.5 =3+05 (-3.6+9) 3t7 7.5)400 り、th ftツ) (-1-2 K(47 3x05) 4.t0 c 4.812 2 y)4.892

stp y h ,y,) O 3A 4.S92t 0.5 (-12 x 4.9 - y)449 I+0 S-/.5 -(.2t 3t Le3e Givan analytul casluton is 0.3x0.5 u312 Kos-4.072 3 (o.S) 7o0.3 ) 4. 322 43 Abrolute enoy le ahyoute 3 3 O 4072 O.622 4.92 4. 322 o.570 4.549 4.169 O 380

Part 3: MATLAB CODE

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clear all
clc;
format short
f=@(t,y)(-1.2*y+7*exp(-0.3*t));   
h=0.5; % arbitrary h value you may change any value accordingly result will be displayed
t = 0:h:1.5;
y = zeros(1,length(t));
y(1)=3; % index has been taken as i instead of 0

for n=1:(length(t)-1)
k1=h*f(t(n),y(n));
k2=h*f(t(n)+h,y(n)+k1);
  
y(n+1)=y(n)+(1/2)*(k1+k2);
end

t_y=[t' y'] % Solution of ODE at t=0 to 1.5

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OUTPUT for h=0.5 (you may change h)

t_y =

0 3.0000
0.5000 3.9462
1.0000 4.1877
1.5000 4.0633