How to get two solution of this single equation

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Kamran
Kamran 2017-8-23
编辑： DGM 2024-1-24
%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
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Torsten
Torsten 2017-8-23
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.

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回答（1 个）

MOSLI KARIM
MOSLI KARIM 2023-11-27
编辑：DGM 2024-1-24
Ran in:
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

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