How to use ode45 on a second-order ODE, but without a function file (just one script)?

I am given the following second-order ODE and I want to use ode45 and by not creating a function file.

I am given the equation above, a timespan t, and initial y and dy values.

I first set up my dydt into a matrix [y1' y2']"

dydt = [y(2); eps*((y(1)^2) - 1)*y(2) - y(1)]

func= dydt(t,y)

and tried to evaluate such ode45(@func, t, [y0, dy0]), but it gives me an error

I also tried setting it up as just one function:

f = @(y1,y2) eps*((y1^2) - 1)*y2 - y1;

Here, it gives me the error that I am using the same function, which was previously used as a variable:

ode45(@f, t, [y0, dy0])

Help please?

Answer 1

[Star Strider]

You are almost there!

Try this:

dydt = @(t,y) [y(2); eps*((y(1)^2) - 1)*y(2) - y(1)];

ic = [1 1]*eps;
tspan = [0 10];

[T,Y] = ode45(dydt, tspan, ic);

figure
plot(T,Y)
grid

That worked when I tried it. Change the initial conditions vector ‘ic’ to match your initial conditions, and ‘tspan’ to get the result you want.

[EB]

Awesome, thank you!

Can I ask for an explanation on what the outputs T and Y are and why there are two output values for Y? Are they just the y and y' values with the tspan I have it set calculating for?

[Star Strider]

My pleasure!

Of course!

The MATLAB ODE solvers calculate values for the independent variable (usually time, so ‘T’ here) and dependent variable vector or matrix (here ‘Y’).

‘Are they just the y and y' values with the tspan I have it set calculating for?’

Yes! The ‘y'’ value is the first column of ‘Y’, and ‘y’ is the second column, corresponding to the rows of your ‘dydt’ function.

[EB]

I see, that makes sense. Thanks again!!

[Star Strider]

As always, my pleasure!