How to get two solution of this single equation

 %The range for one solution 0-20 and for seconde 40-120.

 function My
  xa = 0; xb = 20;
solinit = bvpinit(linspace(xa,xb,1000),[1 0 0]);
sol = bvp4c(@rhs_bvp,@bc_bvp,solinit);
xint = linspace(xa,xb,1000);
Sxint = deval(sol,xint);
plot(x,Sxint(1,:));

function rhs=rhs_bvp(x,y)
rhs=[y(2); y(3) ; y(2)^2-y(1)*y(3)];
end
  function bc=bc_bvp(ya,yb)
      bc=[ya(1)-2; ya(2)+1-ya(3)+ya(4); yb(2)];
  end
end

[KSSV]

solinit is a structure.....you cannot give this as input to the functions @rhs_bvp,@bc_bvp

[Torsten]

I don't understand your question.

By the way: ya(4) does not exist.

Best wishes

Torsten.

[Kamran]

@Torsten, Thanks for your time. I've to use the solution generated by this

y(2)^2-y(1)*y(3)

inplace of

   ya(4)

at x=0. so what should I do for that?

 Actually, I'm solving the attached equations with bvp4c.

[Torsten]

Then, instead of ya(4), you will have to use ya(2)^2-ya(1)*ya(3).

By the way: K is equal to 0 in your example code, not equal to 1.

Best wishes

Torsten.

[Kamran]

Thanks, I solved it.

Answer 1

[MOSLI KARIM]

My

%The range for one solution 0-20 and for seconde 40-120.
function My
xa = 0; xb = 20;
xint = linspace(xa,xb,1000);
solinit = bvpinit(linspace(xa,xb,1000),[1 0 0]);
sol = bvp4c(@rhs_bvp,@bc_bvp,solinit);
Sxint = deval(sol,xint);
plot(xint,Sxint(1,:));

    function rhs=rhs_bvp(~,y)
        rhs=[y(2)
            y(3) 
            y(2)^2-y(1)*y(3)];
    end
    function bc=bc_bvp(ya,yb)
        bc=[ya(1)-2; ya(2)+1-ya(3); yb(2)];
    end
end