dirfield(g,-2:0.2:2,-2:0.2:2)
If you want to use a function several times it is convenient to define it as a so-called inline function:
f1 = inline('sin(x)*x','x')
defines the function f1(x)=sin(x)*x. Note that the arguments of
inline must be strings (not symbolic expressions). You can
then use the function f1 in expressions you type in.
You can also define inline functions of several variables:
g1 = inline('x*y+sin(x)','x','y')
defines the function g1(x,y)=x*y+sin(x) of two variables.
First download the file dirfield.m
and put it in the same directory as your other m-files for the homework.
Define an inline function g of
two variables t, y corresponding to the right hand
side of the differential equation y'(t) =
g(t,y(t)). E.g., for the differential
equation y'(t) = t y2 define
g = inline('t*y^2','t','y')
You have to use
inline(...,'t','y'), even if
t or y does not occur in your formula.
To plot the direction field for t going from t0 to t1 with a
spacing of dt and y going from y0 to y1 with a spacing of dy use
dirfield(g,t0:dt:t1,y0:dy:y1). E.g., for
t and y between -2 and 2 with a spacing of 0.2
type
dirfield(g,-2:0.2:2,-2:0.2:2)
First define the inline function
g corresponding to the
right hand side of the differential equation y'(t) =
g(t,y(t)). E.g., for the differential
equation y'(t) = t y2 define
g = inline('t*y^2','t','y')
To plot the numerical solution of an initial value
problem: For the initial condition y(t0)=y0 you can plot the
solution for t going from t0 to t1 using
ode45(g,[t0,t1],y0).
Example: To plot the solution of the initial value problem y'(t) = t y2, y(-2)=1 in the interval [-2,2] use
ode45(g,[-2,2],1)
The circles mark the values which were actually computed (the points are
chosen by Matlab to optimize accuracy and efficiency). You can obtain vectors
ts and ys with the coordinates of these points using
[ts,ys] = ode45(g,[t0,t1],y0). You can then plot
the solution using plot(ts,ys) (this is a way to
obtain a plot without the circles).
To combine plots of the direction field and several
solution curves use the commands hold
on and hold off: After obtaining
the first plot type hold on, then all subsequent commands plot in
the same window. After the last plot command type hold off.
Example: Plot the direction field and the 13 solution curves with the initial conditions y(-2) = -0.4, -0.2, ..., 1.8, 2:
dirfield(g,-2:0.2:2,-2:0.2:2) hold on for y0=-0.4:0.2:2 [ts,ys] = ode45(g,[-2,2],y0); plot(ts,ys) end hold off
To obtain numerical values of the solution at certain t
values: You can specify a vector tv of t values and
use [ts,ys] = ode45(g,tv,y0). The first element of
the vector tv is the initial t value; the vector tv
must have at least 3 elements. E.g., to obtain the solution with the initial
condition y(-2)=1 at t = -2, -1.5, ..., 1.5, 2 and display the results
as a table with two columns, use
[ts,ys]=ode45(g,-2:0.5:2,1);
[ts,ys]
To obtain the numerical value of the solution at the final
t-value use ys(end) .
sol = dsolve('Dy=t*y^2','t')
The last argument 't' is the name of the independent variable.
Do not type y(t) instead of y.
If Matlab can't find a solution it will return an empty symbol. If Matlab finds several solutions it returns a vector of solutions.
Sometimes Matlab can't find an explicit solution, but returns the solution in implicit form. E.g.,
dsolve('Dy=1/(y-exp(y))','t')returns
t-1/2*y^2+exp(y)+C1=0Unfortunately Matlab cannot handle initial conditions in this case. You can use
ezcontour('t-1/2*y^2+exp(y)',[-4 4 -3 3])to plot several solution curves for t in [-4,4], y in [-3,3]. You can useezplot('t-1/2*y^2+exp(y)-1',[-4 4 -3 3])to plot only the curve where t-1/2*y^2+exp(y)=1.
The solution will contain a constant C1. You can substitute
values for the constant using
subs(sol,'C1',value). E.g., to set C1
to 5 and plot this solution for t=-2 to 2 use
ezplot( subs(sol,'C1',5) , [-2 2] )
To solve an initial value problem additionally specify an initial condition:
sol = dsolve('Dy=t*y^2','y(-2)=1','t')
To plot the solution use
ezplot(sol,[t0,t1]). Here is an example for
plotting the 13 solution curves with the initial conditions y(-2) =
-0.4, -0.2, ..., 1.8, 2:
sol = dsolve('Dy=t*y^2','y(-2)=y0','t') for y0=-0.4:0.2:2 ezplot( subs(sol,'y0',y0) , [-2 2]) hold on end hold off axis tight
To obtain numerical values at one or more t
values use subs(sol,'t',tval) and
double (or vpa for
more digits):
sol = dsolve('Dy=t*y^2','y(-2)=1','t')
This gives a numerical value of the solution at t=0.5:
double( subs(sol,'t',0.5) )
This computes numerical values of the solution at t=-2, -1.5, ..., 2 and displays the result as a table with two columns:
tval = (-2:0.5:2)'; % column vector with t-values
yval = double( subs(sol,'t',tval) )% column vector with y-values
[tval,yval] % display 2 columns together
(continued in Using Matlab for Higher Order ODEs and Systems of ODEs)
Tobias von Petersdorff
