Solving series differential equation use ode45

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Abert
Abert 2020-2-25
评论： Star Strider 2020-2-26
i need to solve the following equations in matlab:
and my codes looks like this :
%initial condition
y0=[0;0;0];
%calling ode function
BO=[0.1 0.2 0.3 0.5 0.8 1 1.5 2];
for i=1:8
bo=BO(i);
[s,y]=ode45(@(s,y) lapalce(s,y,bo),[0 3],y0);
plot(s,y(1));
hold on
end
function dydt=lapalce(s,y,bo)
dydt=[2-bo*y(2)-((sin(y(1))/y(3)));
sin(y(1));
cos(y(1))];
end
matlab is giving me answers like
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
NaN NaN NaN
can anyone tell me what went wrong?

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采纳的回答

Star Strider
Star Strider 2020-2-25
Yes!
The problem is:
y0=[0;0;0];
so:
sin(y(1))/y(3)
will be NaN because sin(0)=0, and 0/0 (and Inf/Inf) result in NaN values.
Settimg ‘y0’ to very small values instead:
y0=[0;0;0]+1E-12;
may give you useful output.
  2 个评论
Abert
Abert 2020-2-26
thank you, i am getting some results now, though it is different than what i thought to be. Need to check my equations.
Star Strider
Star Strider 2020-2-26
As always, my pleasure!
Having all zero initial conditions (or initial estimates in other contexts and applications) is likely not the best approach.

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