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

3 views (last 30 days)
ssmith on 18 Nov 2021
Commented: ssmith on 23 Nov 2021
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)

Cris LaPierre on 19 Nov 2021
Edited: Cris LaPierre on 19 Nov 2021
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.
##### 3 CommentsShow 1 older commentHide 1 older comment
Cris LaPierre on 23 Nov 2021
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 on 23 Nov 2021
@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)

### Categories

Find more on Programming in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!