MATH 267
How to solve ODE systems in MATLAB
MATLAB contains very easy to use commands for numerical solutions of first
order ODE systems.
>> [t,y]=ode45(F,tspan,y0) returns the solution of the initial value problem
for the system defined by F on the interval defined in tspan with initial conditions
defined by y0. If the system has n unknown component functions then the output
array y contains n columns, one for each component.
The inline command can be used to define F as in the case of the single ODE.
(Those of you who know how to use m-files to define functions in MATLAB may do
that instead. See Chapter 8 of the supplementary MATLAB text for more details.)
Suppose for example that the problem of interest is
x01 = tx2 + sin (x1 x2 )
x1 (1) = 3
x02 = x1 e−tx2
x2 (1) = −2
This means that the system may be written in vector form as x0 = F(t, x) where
tx2 + sin (x1 x2 )
F(t, x) =
x1 e−tx2
A MATLAB command which will define this vector function is
>> F=inline(’[t*x(2)+sin(x(1)*x(2));x(1)*exp(-t*x(2))]’,’t’,’x’)
The input tspan defines the interval of t values on which the solution is to be
found, as in the single equation case, e.g.
>> tspan=[1 4]
will cause the solution to be computed for 1 ≤ t ≤ 4, with initial conditions at t = 1
given by y0, which in this case should be
>> y0=[3;-2]
(Why the semicolon in y0? It forces y0 to be a column vector. Without the semicolon, as in the definition of tspan, you are defining a row vector. The semicolon
also occurs in the definition of F because it needs to be a column vector, but tspan
can be either a row or column vector.)
To plot the solution, give the command
>> plot(t,y(:,1));shg
which will create a plot of the first component of the solution. Replace 1 by 2 to get
the second component. (shg shows the graph window, if it isn’t already visible).
You can get both curves on the same plot with the command
>> plot(t,y(:,1),t,y(:,2));shg
To display the solution in the phase plane
>> plot(y(:,1),y(:,2));shg
Homework
1. Obtain the numerical solution of the problem stated on the reverse side for
1 ≤ t ≤ 4 by entering all of the commands there in an appropriate order. Plot both
components of the solution on the same axes.
2. Transform the problem
t
y 00 = 3y − sin (y 0 − )
2
0 < t < 2 y(0) = 1
y 0 (0) = −2
into an equivalent problem for a first order system and solve numerically using
ode45. Plot the solution. (Remember the solution is a single function in this case,
not two functions.)
3. Use ode45 to solve
−1 −4
x =
x
5 −1
0
2
x(0) =
1
Compute the solution for 0 < t < 10 and plot the result in the phase plane.
You do not need to print out columns of numbers in any of these problems. Turn
in plot only for 1 and 3, and for problem 2 turn in the plot and write out the first
order system you used. Refer to the first MATLAB assignment if needed to remind
yourself how to get a plot printed.
4. Find the exact solution of problem 3 by elimination as follows: Solve the first
equation for x2 in terms of x01 , x1 , differentiate to obtain x02 in terms of x001 , x01 , and
then substitute for both x02 , x2 in the second equation. You should end up with a
single second order constant coefficient equation for x1 , which may be solved by
the method of Chapter 4. Use the first equation again to obtain x2 and then fit the
initial conditions.