what can i further do in my this code of linear shooting

function[]=LS(x,y) clc % Dy is the representation of y' %D2y is the representation od y'' D2y=inline('-(2/x)*Dy + (2/x^2)*y + sin(ln(x))/x^2','x','y','Dy') p=inline('-(2/x)','x'); p1=p; q=inline('(2/x^2)','x'); q1=q; r=inline('sin(log(x))/x^2','x'); r1=r; a=input('enter the lower limit of x: '); b=input('enter the upper limit of x: '); N=input('enter the integer N: '); y0=input('enter the lower boundary value of y: '); y1=input('enter the upper boundary value of y: '); fprintf('The system of linear Differential Equations the become\n') yA=inline('-(2/x)*Dy + (2/x^2)*y + sin(ln(x))/x^2','x','y','Dy') fprintf('with the conditions y(%1g)=%1g and Dy(%1g)=0\n',a,y0,a) f1=yA; yB=inline('-(2/x)*Dy + (2/x^2)*y','x','y','Dy') fprintf('with the conditions y(%1g)=0 and Dy(%1g)=1\n',a,a) f2=yB; h=(b-a)/N; u1=y0; u2=0; v1=0; v2=1; w1=y0; fprintf('\n x u1 v1 w1\n'); fprintf('%10g %10g %10g %10g\n',a,u1,v1,w1); for i=0:N-1; x=a+i*h; K11=h*(u2); K12=h*(((p1(x)*u2))+(q1(x)*u1)+r1(x)); K21=h*(u2+(1/2)*K12); K22=h*((p1(x+h/2)*(u2+(1/2)*K12)+q(x+h/2)*(u1+(1/2)*K11)+r(x+h/2))); K31=h*(u2+(1/2)*K22); K32=h*((p1(x+h/2)*(u2+(1/2)*K22)+q(x+h/2)*(u1+(1/2)*K21)+r(x+h/2))); K41=h*(u2+K32); K42=h*((p(x+h)*(u2+K32)+(q(x+h)*(u1+K31))+r(x+h))); u1=u1+(1/6)*(K11+2*K21+2*K31+K41); u2=u2+(1/6)*(K12+2*K22+2*K32+K42);

    k11=h*(v2);
    k12=h*(((p1(x)*v2))+(q1(x)*v1));
    k21=h*(v2+(1/2)*k12);
    k22=h*((p1(x+h/2)*(v2+(1/2)*k12))+(q(x+h/2)*(v1+(1/2)*k11)));
    k31=h*(v2+(1/2)*k22);
    k32=h*((p1(x+h/2)*(v2+(1/2)*k22))+(q(x+h/2)*(v1+(1/2)*k21)));
    k41=h*(v2+k32);
    k42=h*((p(x+h)*(v2+k32))+(q(x+h)*(v1+k31)));
    v1=v1+(1/6)*(k11+2*k21+2*k31+k41);
    v2=v2+(1/6)*(k12+2*k22+2*k32+k42); 
    fprintf('%10g %10g %10g\n',x+h,u1,v1);
end

[C.J. Harris]

The first thing I'd suggest doing is formatting it better.

[Jan]

In the current form your code looks very ugly. Please format your code using the "{} Code" button. Follow the link "Markup" on this page also.

[Muhammad Usman]

function[]=LS(x,y)

clc

% Dy is the representation of y'

%D2y is the representation od y''

D2y=inline('-(2/x)*Dy + (2/x^2)*y + sin(ln(x))/x^2','x','y','Dy')

p=inline('-(2/x)','x');

p1=p;

q=inline('(2/x^2)','x');

q1=q;

r=inline('sin(log(x))/x^2','x');

r1=r;

a=input('enter the lower limit of x: ');

b=input('enter the upper limit of x: ');

N=input('enter the integer N: ');

y0=input('enter the lower boundary value of y: ');

y1=input('enter the upper boundary value of y: ');

fprintf('The system of linear Differential Equations the become\n')

yA=inline('-(2/x)*Dy + (2/x^2)*y + sin(ln(x))/x^2','x','y','Dy')

fprintf('with the conditions y(%1g)=%1g and Dy(%1g)=0\n',a,y0,a)

f1=yA;

yB=inline('-(2/x)*Dy + (2/x^2)*y','x','y','Dy')

fprintf('with the conditions y(%1g)=0 and Dy(%1g)=1\n',a,a)

f2=yB;

h=(b-a)/N;

u1=y0;

u2=0;

v1=0;

v2=1;

w1=y0;

fprintf('\n x u1 v1 w1\n');

fprintf('%10g %10g %10g %10g\n',a,u1,v1,w1);

for i=0:N-1;

x=a+i*h;

K(1,1)=h*(u2);

K(1,2)=h*(((p1(x)*u2))+(q1(x)*u1)+r1(x));

K(2,1)=h*(u2+(1/2)*K(1,2));

K(2,2)=h*((p1(x+h/2)*(u2+(1/2)*K(1,2))+q(x+h/2)*(u1+(1/2)*K(1,1))+r(x+h/2)));

K(3,1)=h*(u2+(1/2)*K(2,2));

K(3,2)=h*((p1(x+h/2)*(u2+(1/2)*K(2,2))+q(x+h/2)*(u1+(1/2)*K(2,1))+r(x+h/2)));

K(4,1)=h*(u2+K(3,2));

K(4,2)=h*((p(x+h)*(u2+K(3,2))+(q(x+h)*(u1+K(3,1)))+r(x+h)));

u1=u1+(1/6)*(K(1,1)+2*K(2,1)+2*K(3,1)+K(4,1));

u2=u2+(1/6)*(K(1,2)+2*K(2,2)+2*K(3,2)+K(4,2));

k(1,1)=h*(v2);

k(1,2)=h*(((p1(x)*v2))+(q1(x)*v1));

k(2,1)=h*(v2+(1/2)*k(1,2));

k(2,2)=h*((p1(x+h/2)*(v2+(1/2)*k(1,2)))+(q(x+h/2)*(v1+(1/2)*k(1,1))));

k(3,1)=h*(v2+(1/2)*k(2,2));

k(3,2)=h*((p1(x+h/2)*(v2+(1/2)*k(2,2)))+(q(x+h/2)*(v1+(1/2)*k(2,1))));

k(4,1)=h*(v2+k(3,2));

k(4,2)=h*((p(x+h)*(v2+k(3,2)))+(q(x+h)*(v1+k(3,1))));

v1=v1+(1/6)*(k(1,1)+2*k(2,1)+2*k(3,1)+k(4,1));

v2=v2+(1/6)*(k(1,2)+2*k(2,2)+2*k(3,2)+k(4,2));

fprintf('%10g %10g %10g\n',x+h,u1,v1);

end

[Muhammad Usman]

now guide me please to generate the solutions...

Answer 1

[Jan]

Do not use "k24" if you want to access an array - "k(2, 4)" is smarter, faster and the musltiplication by "h" can be performed for all elements simultaneoulsly with "h*k".