Question

please slove this question in MATLAB and explain clearly about it. Many thanks.

Question 3. Given the following boundary-value problem Use the finite difference method to solve the problem with h 0.02. Plot your results.

Answer 1

Matlab code for solving Finite difference method ode clear all close all %creating tridiagonal matrix dt-0.02; a_ @ ( tt ) (2-(8*tttl ) * dt ) ; b-e (tt ) (2칙dt "2 ) * (-(16+tt."2))-4); c-@ ( tt) tridiagonal matrix of size n ( (2-0)/dx)-1; tall initial conditions t_in-0; t fnl-1; y in-l; y_end-0; t-t_in:dt:t_fnl; number of steps n = ( ( t-fnl-t-in ) /dt )-1 ; Tridiagonal matrix %Arfull( gallery ( .tridiag. ,n, a,b,c) ) ; A-zeros(n,n); for i-2:n-1 end A(n,n-1)-a(t(n)) A(n,n)-b (t (n) for i-2:n end bb (n,l) bb (n,1)-c(t(n))*y_end; %solving using backslash operator tall u values y val-A\bb; for i-2: n+1 yy(i)-y val(i-1); end yy (n+2)-y_end; all x value xx-t in:dt:t_fnl; plot (xx, yy, Linewidth',2) xlabelt) ylabel('y(t) title( 'y(t) vs. t plot using finite difference method for h-0.02') fprintf( The solution vector for y n) disp (yy)

,,움各各움움各各움움各움움움各各움움各움 End of Code 各各各各各各움음움움움움움움움움움움움음 The solution vector for y- 0.911117026058629 0.829767525825777 0.755531542134532 0.687995159093996 0.626751598943911 0.571402257103799 0.521557666109205 0.476838380334544 0.436875774 704702 0.401312751967667 0.36980435450926 0.342018278109691 0.31763528644253 0.296349526473382 0.2778687462042 0.261914417408313 0. 24822176709266 0. 23653972239254 0.226630774439205 0.218270767434019 0.211248619 710958 0.205365983971149 0.200436854131835 0.196287126353276 0.192754121798932 0.1896860 78557292 0.186941619920211 0.184389205886065 0.181906574350854 0.179380177981353 0.176704622246375 0. 17378210952958 0.170521893673858 0.166839748725893 0.162657455071801 0.157902305590949 0.152506633914179 0.146407366361947 0. 13954559866332 0.131866198122599 0.123317431509756 0.11385061860532 0.103419811030566 0.0919814957391407 0.0794943223353304 0.0659188532147165 0. 0512173353924357 0.0353534927893017 0. 0182923376833267

y(t) vs. t plot using finite difference method for h-0.02 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.1 02 034 0.5 06 07 00.91 Published with MATLABR R2018a

%%Matlab code for solving Finite difference method ode
clear all
close all
%creating tridiagonal matrix
dt=0.02;
a=@(tt) (2-(8*tt+1)*dt); b=@(tt) (2*(dt^2)*(-(16+tt.^2))-4); c=@(tt) (2+(8*tt+1)*dt);
r=@(tt) (-12)*tt.^2;
%tridiagonal matrix of size n=((2-0)/dx)-1;
%all initial conditions
t_in=0; t_fnl=1;
y_in=1; y_end=0;

t=t_in:dt:t_fnl;
%number of steps
n = ((t_fnl-t_in)/dt)-1;
%Tridiagonal matrix
%A=full(gallery('tridiag',n,a,b,c));
A=zeros(n,n);
for i=2:n-1
    A(i,i)=b(t(i));
    A(i,i+1)=c(t(i));
    A(i,i-1)=a(t(i));

end
A(1,1)=b(t(1));
A(1,2)=c(t(1));
A(n,n-1)=a(t(n));
A(n,n)=b(t(n));

for i=2:n
    bb(i,1)=(2*dt^2*r(t(i)));
end
bb(1,1)=bb(1,1)-a(t(1))*y_in;
bb(n,1)=bb(n,1)-c(t(n))*y_end;
%solving using backslash operator
%all u values
y_val=A\bb;
yy(1)=y_in;
for i=2:n+1
    yy(i)=y_val(i-1);
end
yy(n+2)=y_end;
%all x value
xx=t_in:dt:t_fnl;
plot(xx,yy,'Linewidth',2 )
xlabel('t')
ylabel('y(t)')
title('y(t) vs. t plot using finite difference method for h=0.02')

fprintf('The solution vector for y=\n')
disp(yy')
%%%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%%%