# ODE45 help varying values

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john doe on 7 Apr 2018
So I have a function and I am trying to solve it via ODE45. It works perfectly and gives me my solution and plots it on a nice little graph. What I am struggling to understand is that when I change my parameter value, a different amount of Y values are produced. For example, my code is as follows;
z=0;
f = [1 2 3]
ft =[1 2 3];
g = [4 5 6];
gt = [1 2 3];
Time = [0 30];
IC = 2.5;
for a = [1:3]
for b = [1:3]
Q = [a b];
[t,y] = ode45(@(t,y) Test(t,y,Q,ft,f,gt,g), Time, IC);
end
y=y'
z=z+1;
out(z,:) = [a b y]
end
with an ODE file;
function dydt = Test(t,y,Q,ft,f,gt,g)
f = interp1(ft, f, t,'linear', 'extrap');
g = interp1(gt, g, t, 'linear', 'extrap');
dydt = Q(1)*f+Q(2)*g*y;
end
Now if for example I let a=1 and b=1, then I get 1845 Y values. If however I set a=1 and b=2 then I get 2427 Y values. I really do not understand how simply changing the parameter gives a different 'amount' of solutions. It is making it very difficult to store my solutions in a matrix since I keep getting an error about them not matching up.

Razvan Carbunescu on 9 Apr 2018
The height of the answer is the number of timesteps needed to compute the solution. The a=1, b=2 solution takes more steps to compute to the same T=30 time.
One way to make them match up is to interpolate the solution [t,y] to a common time interval
[t1,y1] = ode45(...)
[t2,y2] = ode45(...)
y2i = interp1(t2,y2,t1);