MATH 308 Sheet 1
Some syntax trouble spots:
multiplication
3*t for 3t
powers
x∧2 for x2
number π
Pi
Greek letter π
pi
sin(x)
abs(x)
cos(x)
sqrt(x)
tan(x)
ln(x)
exp(x)
for
for
for
for
for
for
for
sin x
|x|
cos x
√
x
tan x
ln x
ex
Maple can be used to plot direction fields and solution curves. You must load the DEtools
package once on each worksheet:
> with(DEtools):
Note the colon will suppress any output from Maple, whereas a semicolon will not.
dy
Example 1.
= −y.
dx
Assign the differential equation the name de for easy handling and to avoid trouble, always
type the dependent variable y as y(x).
> de:=diff(y(x),x)=-y(x);
> DEplot(de,y(x),x=-3..3,y=-3..3);
To plot the direction field and solution curves, for example the solutions satisfying y(1) = 2,
y(−1) = −1 and y(1) = 1, proceed as follows:
> inits:=[ [1,2],[-1,-1],[1,1] ];
Here we’re telling Maple the initial conditions in the appropriate form. Always be sure to
enclose the list in square brackets.
> DEplot(de,y(x),x=-3..3,inits,y=-3..3);
You might need to play around with the x and y plot ranges to get a good plot.
If you just want a plot of the solution curves, include the arrows=none option:
> DEplot(de,y(x),x=-3..3,inits,y=-3..3,arrows=none);
NOTES
1. For good printouts, include the option linecolor=black to make the solution curves
black.
2. If your solution curves appear jagged, include the option stepsize=h, where you choose
h by trial and error to get a good plot. For instance, try .1, .05, .01 etc. Please note
on exam, your solution will lose credit if your solution curves appear jagged. Place the
option after the y range.
3. To resize Maple’s plots, click on the graph and drag the corners with the mouse.
4. Use the initial conditions to help you pick the x and y plot ranges. For instance, if
y(−3) = −1, use x=-6..0, y=-4..2 as a starting point and play around from there if
necessary.
5. The command restart: will clear all values of variables. It’s a good thing to try when
things go wrong.
6. To type text in a Maple worksheet, hit the button with the T on it. To restore the
Maple prompt, hit the button with the [ > on it.
dy
Example 2.
= sin(y). Plot the direction field using Maple. What happens to the
dx
solution satisfying
1. y(0) = 1 as x → ∞.
2. y(2) = −2 as x → ∞.
3. y(0) = 7 as x → ∞.
Example 3. The population p(t) in thousands of a certain species satisfies the differential
dp
= 3p − 2p2 . Use Maple to sketch the direction field and use it to answer the
equation
dt
following questions.
1. If the initial population is 2000 individuals (i.e., p(0) = 2), what is the limiting population?
2. If the initial population is 500 individuals, what is the limiting population?
3. Can a population of 3000 individuals ever decline to 500 individuals?
Example 4. For a bar magnet, the magnetic field lines satisfy the differential equation
dy
3xy
= 2
. Plot the direction field.
dx
2x − y 2
We have to iterate the algorithm that realizes an Euler’s method to construct approximations
to the solution of the initial value problem for first-order differential equation:
dy
= f (x, y),
y(x0 ) = y0 .
dx
As you know, an Euler’s method can be summarized by the recursive formulas
xn+1 := xn + h,
yn+1 := yn + f (xn , yn ),
n = 0, 1, 2, . . .
Lets construct numerical approximations to the initial value problem
1
y
dy
= 2 − − y2,
dx
x
x
y(1) = 1
on the interval 1 < x < 2.
To iterate this algorithm, we use the for k from start to finish do..od construction. The
following sequence of Maple commands performs ten iterations of this procedure with step
h = 0.1 and therefore computes approximate values at the points x = 1, 1.1, 1.2, ...1.9, 2.0.
These values are contained in a sequence of points named eseq.
> f:=(x,y) -> 1/xˆ2-y/x-yˆ2;inits:=y(1)=1;
1
y
− − y2
2
x
x
inits := y(1) = 1
f := (x, y) →
Notice, that the function above is an arrow-defined function an not a Maple expression.
> x:=1:y:=1:h:=0.1: # initialize x and y an the step size
> eseq:=[x,y]; # input the initial conditions into eseq
eseq := [1, 1]
> for i from 1 to 10 do
y:=evalf(y+h*f(x,y)): # compute the new value of y
x:=x+h: # update the new value of x
eseq:=eseq,[x,y]: # ad the new point
od:
> x:=’x’:y:=’y’:h:=’h’:
To display the contest of the solution values in eseq, type the variable name.
> eseq;
[1, 1], [1.1, .9], [1.2, .8198264463], [1.3, .7537404800], [1.4, .6981195696], [1.5, .6505372009],
[1.6, .6092926336], [1.7, .5731505927], [1.8, .5411877679], [1.9, .5126975583], [2.0, .4871284287]
To plot the approximation between x = 1 and x = 2, we could use plot([eseq]);.
1.0
0.9
0.8
0.7
0.6
0.5
1.0
1.2
1.4
1.6
1.8
2.0
1. Use Euler’s method to find approximations to the solution of the initial value problem
y ′ = 1 − sin y,
y(0) = 0,
taking 10 steps.
2. Use Euler’s method to find approximations to the solution of the initial value problem
y ′ = 1 + x2 ,
taking 10 steps.
y(0) = 0,