I have initial value problem that I have to write the right side into a vector function

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ssmith
ssmith 2021-11-18
评论： ssmith 2021-11-23
I have DEs that I need to write into a vector function and use
ode45
to approximate the solution for the initial value problem on the interval 0 < t < 12. This is what I have so far
w = @(t,y) 5 + (z/5) - ((4*y)/(20+3*t))
w1 = @(t,z) ((4*y)/(20+3*t)) - (2*z)/5
[t,y] = ode45(w, [0, 12], 0)
[t,z] = ode45(w1, [0, 12], 20)

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回答（1 个）

Cris LaPierre
Cris LaPierre 2021-11-19
编辑：Cris LaPierre 2021-11-19
The hints are in the instructions. You are not solving the problem the way you are asked to.
Because your equations are coupled, you have to solve them simultaneously. To do that, create an odefun that solves both equations. The input (initial conditions) and output (solution of both equations) are vectors. The output must be a column vector.
See this example (the odefun with 2 equations).
Adapting it to your case, you could designate y(1) to be y in your equations, and y(2) to be z. As long as the order is consistant and matches your y0 input, it doesn't matter which one is y and which one is z.
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Cris LaPierre
Cris LaPierre 2021-11-23
Ran in:
Well, let's give the credit to @William Rose, as he is the one who came up with this code.
Just run it. What is it you want to do with the results? Perhaps a plot? See below.
y0 = [0;20];
tspan = [0,12];
[t,y] = ode45(@odefun,tspan,y0);
plot(t,y)
legend('y','z')
function dydt = odefun(t,y)
dydt = zeros (2,1);
dydt(1) = 5 + y(2)/5 - 4*y(1)/(20+3*t);
dydt(2) = 4*y(1)/(20+3*t) - 2*y(2)/5;
end
ssmith
ssmith 2021-11-23
@Cris LaPierre Yes, I am trying to approximate the solution for the initial value equation on the interval of 0 <= t <= 12 and plot the resulting y(t) and z(t)

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