Reputation: 159
I need to solve the following
-cos(y)y''+sin(y)y'^2+sin(y)=0, y'(0)=y'(1)=0, such that y=y(t)
I find it hard to solve because of the term y'^2
and also the boundary conditions.
Upvotes: 0
Views: 601
Reputation: 3014
Here is the Scilab code for your bvp
-cos(y)y''+sin(y)y'^2+sin(y)=0, y'(0)=y'(1)=0, y(0)=0, y(1)=1.5
but with different boundary condition not giving the trivial solution. You have first to write y'' as a function of y and y'. The function fsub
computes y'' as a function of u=[y,y']
function ysecond=fsub(x,u)
y=u(1);
yprime=u(2);
ysecond = sin(y)/cos(y)*(1+yprime^2);
end
function g=gsub(i, u),
y=u(1);
select i
case 1 then // x=zeta(1)=0
g = y // y(0)=0
case 2 then // x=zeta(2)=1
g = y-1.5 // y(1)=1.5
end
end
N=1;
m=2;
x_low=0
x_up=1;
xpoints=linspace(0,1,100);
zeta=[0,1];
u = bvodeS(xpoints,m,N,x_low,x_up,fsub,gsub,zeta)
plot(xpoints,u(1,:))
Upvotes: 3