# Saving each vector of an ode45 vector solution in a matrix.

조회 수: 2(최근 30일)
Gabriel Venter 2021년 4월 26일
댓글: J. Alex Lee 2021년 4월 27일
I have the following code, where I am using a different value of a parameter G in my ODE system (solving it w/ ode45) and want to plot each solution on the same graph. It isnt the same as IC's, as these are the same but the parameter varies. Currently I have:
for i=1:20
G=i/20 %Saving that iterates value of G
[phi,y]=ode45(@(phi,y) odefun(phi,y,G), timerange, IC);
z(i,;)=y %
end
then I want to save all these y's without creating 20 vectors. Is there a way to save each vector y as a column of a matrix? I am very lost and any direction would be very helpful, I cannot find a thread or anythign on the helpcenter.
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Gabriel Venter 2021년 4월 27일
I have a 2D autuonomous ODE system here. And I was worried about this timestep problem since ode45 uses a variable timestep and that would be different for different G, so clearly what I was doing was stupid. Thanks!

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### 답변(1개)

J. Alex Lee 2021년 4월 27일
편집: J. Alex Lee 2021년 4월 27일
z = cell(1,20)
for i=1:20
G=i/20 %Saving that iterates value of G
[phi,y]=ode45(@(phi,y) odefun(phi,y,G), timerange, IC);
z{i}=y %
end
Or better yet just use the solution structure version of output
for i=20:-1:1
G=i/20 %Saving that iterates value of G
sol(i) = ode45(@(phi,y) odefun(phi,y,G), timerange, IC);
end
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J. Alex Lee 2021년 4월 27일
"20:-1:1" is going backwards, so the pre-allocation of the structure array is implied. Didn't need to do it with the cell array version since it was straightforward to pre-allocate an empty cell array. You could loop backward on that example as well, and wouldn't need to do the explicit pre-allocation.

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