Solving using an ode

I am new to matlab and have not been able to figure out this question. I have to use the function 'ode45' do derive dx/dt=q*x^(s)+w*x.

some info:

0<=t<=20

the initial value of x=0.2.

q=10
w=3
s=[0.60 0.64 0.68 0.72 0.76 0.80]
t1=[5.34 4.25 3.25 2.12 1.96 1.52]

the question asks to solve the equation using ode45 function and use the solution to find the time t it would take for x =10 for every value of s

Here is the code I have, I know it is wrong but i thought I would just have it here.

s=[0.60 0.64 0.68 0.72 0.76 0.80];
t1=[5.34 4.25 3.25 2.12 1.96 1.52];
q=10;
w=3;
x0=0.2;
x1=10;

ode@(t1,x)(q*x^s+w*x);
[t1,x]=ode45(@ode,[0:20],0.2);

Answer 1

[Bjorn Gustavsson]

You want to integrate your ode for a couple of values of s. To do that you can either create one ode-function for all 6 values of s or you could do the ode-integration for one value of s at a time. When it comes to integrating multiple uncoupled ODEs it might take longer time to call ode45 with them all at once because ode45 uses adptive time-steps - and the separate values of s might lead to shorter steps in different regions of your time-variable, on the other hand it might also be quicker. If you do the integration of the odes one-by one you will have to create the variable ode once for each value of s, perhaps in a loop and save the results somehow. Once you've made that work you should also have a look at the way odeNN handles "events" that would make it easier to achieve the goal for you.

HTH

Answer 2

[Alan Stevens]

Here's a "quick and dirty" way to do it, using interpolation to find the time.

s = [0.60, 0.64, 0.68, 0.72, 0.76, 0.80];
x0 = 0.2;
x1 = 10;
tf = zeros(1, numel(s));
tend = 0.4;

for k = 1 : numel(s)
    [t, x] = ode45(@odefn, [0, tend], x0, [], s(k));
    tf(k) = interp1(x, t, x1);
    plot(t, x)
    hold on
end

plot([0, tend], [x1, x1], 'r')
xlabel('t')
ylabel('x')
grid
text(0.01, 20, sprintf('tf = %.2f\n', tf))

function dxdt = odefn(~, x, s)
    q = 10;
    w = 3;
    dxdt = q * x ^ s + w * x;
end

[Alan Stevens]

A slightly cleaner looking version madhan, though sprintf could do with extra spaces for clarity on the graph!.

That's better!