Lab5_Instructions
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Jan 9, 2024
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Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
1
In this lab, we study a differential equation with complex roots and force it with a sinusoidal function. This will
help you master the
Method of
Undetermined Coefficients
. We will first look at the homogeneous DE, and then
add various multiples of the forcing term to illustrate linearity and superposition.
Part A: Homogeneous DE
Question 1:
Consider the initial value problem:
Note this 2
nd
-order equation is linear and homogeneous. We will add a forcing term in part B.
Using
pplane8
, create a phase diagram, for this DE.
a. Proceed
and show the
nullclines
. In MATLAB, nullclines are
Contours
.
Under
Solutions
on the pplane8 Display window, select
Show nullclines
. All solutions that cross a nullcline
will do so at a local max or min. There are two nullclines, one for each of the variables
?
1
and
?
2
.
The respective nullclines are where the derivative of either variable is zero.
Nullcline for
?
1
:
𝑑
𝑑𝑡
?
1
= ?
2
= 0
So this nullcline is the horizontal axis
?
2
= 0
Nullcline for
?
2
:
𝑑
𝑑𝑡
?
2
= − (
17
4
) ∗ ?1 − ?2 = 0
So this nullcline is the line
?
2
= − (
17
4
) ∗ ?1
Make the axes equal so circles look like circles and not ellipses.
We decompose this 2
nd
ODE into two 1
st
ODE to use with
pplane8.
These two equations are:
?
1
′
= ?
2
?
2
′
= −(17/ 4) ∗ ?
1
− ?
2
Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
2
So far, your phase plane should look like this.
b.
Next let's add some solution curves. First, use
Options
and set the
Solution direction
to
Forward
only.
Next, using
Keyboard input
, find the solutions for each of the initial points below. Show the first in
magenta
and second in
blue
. Increase their
LineWidth
to 3.
IC:
(𝟒, 𝟎),
(−𝟒, 𝟎)
Q1: Paste your completed phase plot into the answer template.
Be sure the solution for (4,0) is in
magenta
and that for (
–
4,0) is in
blue
. Be sure the nullclines are visible.
Replace this
sample graph, which has some of the graph hidden with an unrelated image of panther cubs.
Sample
x1 ' = x2
x2 ' = - (17/4) x1 - x2
-5
0
5
x1
-5
-4
-3
-2
-1
0
1
2
3
4
5
x2
Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
3
Question 2:
Find the exact solution to this same initial value problem using
dsolve
.
Tips:
Declare
y(t)
as symbolic first using
syms
. Then within
dsolve
, use
==
to denote equality in equations.
Q2:
Copy the exact solution into the answer template including the code used to find it.
Question 3:
Use the help to review
matlabFunction()
.
Use it to define the functions Y(t) and DY(t), its derivative. These represent the exact solution and its
derivative.
Hint:
If you let
sol
denote the exact solution from
dsolve
,
sol = dsolve( ... )
% Arguments hidden.
then you can make that answer into a function using:
Y = matlabFunction(sol)
Now define DY using
Y
, the
diff()
command and another
matlabFunction()
.
Next evaluate
𝑌(0)
and
𝐷𝑌(0)
to verify both your functions are working.
Of course, we expect
𝑌(0) = 4
and
𝐷𝑌(0) = 0
, since
𝑌(?)
is the solution to the IVP.
Q3: Now evaluate both functions at time
t
=1. Paste in your values for Y(1) and DY(1) below.
This demonstrates you can evaluate the solution and its derivative anywhere!
%% Q2 - Solve the Homogeneous IVP exactly using dsolve.
syms
y(t)
Dy = diff(y,t); D2y = diff(y,t, t);
%
The exact solution is:
?(?) = ⋯
Alert:
High risk for the
MATLAB Exam!
Q3: Answer
>>
𝑌(1)
=
_ _ _ _ _
>>
𝐷𝑌(1)
=
_ _ _ _ _
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Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
4
Q4-5:
Using your exact functions Y(t) and DY(t), and subplot commands, create
component plots
for Y and DY
and their phase plot. Do
not
use
ode45
or any other approximate solver. You already have the exact answers!
Here is the start of the first of the three plots to get you started. We'll cut the plot into six tiles (instead of 4) to
allow more room for the phase plot on the right. We'll draw Y(t) in
magenta
in
tile 1
and its derivative in
red
in
tile 4
. The phase plot on the right will take all four of the tiles 2, 3, 5 and 6.
Now add a plot of DY(t) in tile 4 (bottom left) and add the phase plot on the right using both tiles 2, 3, 5 and 6,
using [2, 3, 5, 6] in your code. Label the axes, give appropriate titles, set the 'FontSize' to 20 and draw DY in
red
and the phase curve in
black
. Mark the starting point (4,0) in the phase plot. Add a legend to the phase plot as
shown. Mark the initial point with a blue circle filled in
green
.
Mark the center of the circle point as a blue
hexagram filled in
yellow
. Here's some sample code.
>> plot(0, 0,
'bh'
,
'MarkerSize'
, 14,
'MarkerFaceColor'
,
'yellow'
)
% Center Point
Questions 4-5: Paste your completed component and phase plots into the answer template for credit.
Replace this sample graph, which has some of the image hidden by amethyst crystals.
figure
subplot(2,3,1)
% tile 1
time = 0: 0.01 : 10;
plot(time, Y(time),
'm'
,
'LineWidth'
,3)
grid
on
hold
on
4
5
6
1
2
3
Sample
Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
5
Part B: Non-Homogeneous DE: Method of Undetermined Coefficients
Consider the
non-homogeneous
initial value problem:
Questions 6-7:
Complete the table below, to find the unique solution matching the initial conditions and the
given forcing function
?(?)
which varies in each row. Simply use
dsolve()
.
The homogeneous part is given, so you just need to add on the particular terms right after the trailing + sign.
If you do not see the blue terms in the solution returned by
dsolve
, try
simplify()
.
Forcing function
𝒇(𝒕)
Guess for Particular
Solution
Unique Solution y(t)
a.
?(?) = 17 + 289?
?? + ?
?(?) = ?
−
𝑡
2
∙ [7 cos 2? −
27
4
sin 2? ]
+
…
b.
?(?) = 100 ?
−2𝑡
? ?
−2𝑡
?(?) = ?
−
𝑡
2
∙ [4 sin 2? ]
+
…
c.
?(?) = 260 cos 2?
? cos 2?
+
? sin 2?
?(?) = −?
−
𝑡
2
∙ [32 sin 2? ]
+
…
d.
?(?) = 16?
−
𝑡
2
cos 2?
+
32 ?
−
𝑡
2
sin 2?
Bump up!
? ?
−𝑡/2
[? cos 2? + ? sin2?]
?(?) = ?
−
𝑡
2
∙ [4 cos 2? + 2 sin 2? ]
+
…
* Grader will randomly pick two to check for correctness.
Question 8:
Let's focus on part (
c)
above. This time, we will not rely on
dsolve
, but will implement the
Method
of Undetermined Coefficients
from scratch! Based on Lecture 4-5, you should expect a particular solution of
the form below, as this would be closed under differentiation.
?
𝑝
(?) = ? ∙ cos(2?)
+
? ∙ sin(2?)
Since
the functions
cos(2?)
and
sin(2?)
are
not
solutions of the homogeneous equation so we will
NOT
need
to
bump it up
! Let's verify our solution has the above form:
Enter the above DE and find the exact solution by substituting our guess for
?
𝑝
(?)
into the original DE, which
now includes the forcing term
?(
?)
= 260
cos
2?
.
Let's solve for
A
and
B
as follows. See next page.
Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
6
This gives:
(8*B + A)*cos(2*t) + (B - 8*A)*sin(2*t) == 260*cos(2*t)
The coefficients of both terms must match on each side:
cos(2t): A + 8B = 260
sin(2t):
-8A + B = 0
so
? = 8?
Plugging
? = 8?
into the first equation we find:
? =
260
65
= 4,
? = 32
Thus, the method of undetermined coefficients leads to the particular solution:
?
𝑝
(?) = 4 ∙ cos(2?)
+
32 ∙ sin(2?)
Questions 9-10: Graph the Non-Homogeneous Solution
Tip:
You can clone most of your plotting code from Questions 4-5 to create the plot for this part. Of course, both
the functions
?(?)
and
?′(?)
have changed. You will earn the remaining two points by pasting in your completed
graph for these questions.
DE:
𝟒?
′′
+ 4?
′
+ 17? =
𝟐𝟔𝟎 𝐜𝐨𝐬(𝟐𝒕)
IC:
?(0) = 4
?
′
(0) = 0
Create a new figure using:
fig3 = figure(3);
% Start with a new figure using ***SIX TILES*** in a 2x3 grid
subplot(2,3,1)
% We'll plot the solution Y(t) in tile #1 in magenta
%% QUESTION 8: The Method of Undetermined Coefficients from SCRATCH
clc
syms
A B t
syms
y(t) a b
Dy = diff(y,t);
D2y = diff(y,t, t);
f = 260 * cos(2*t)
DE = 4* D2y + 4*Dy + 17*y - f == 0
% non-homogeneous differential equation
Y = A * cos(2*t)
+
B * sin(2*t)
% our guess
plug_it_in = subs(DE, y, Y)
% Plug our guess into the DE
eqn = collect(plug_it_in, [cos(2*t), sin(2*t)])
% arrange by similar terms
% MATLAB can find the undetermined coefficients for us
equations = coeffs( lhs(eqn), [cos(t), sin(t)] )
variables = [A, B]
% find and display the undetermined coefficients
[A, B] = solve(equations, variables)
Question 8:
Now that you have found the particular solution, record its derivative for one point.
?
𝑝
′
(?) =
_ _ _ _ _
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Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
7
We will use the same 2x3 grid we used earlier in Question 4-5.
a
. In the first tile, plot the solution
?(?)
using your solution from
dsolve
found earlier. Use
magenta
and a
linewidth of 3. Be sure your forcing function is
?(?) =
260 cos 2?
.
b
. In the fourth tile, plot the solution for its derivative
?′(?)
using your solution from
dsolve
found earlier and
the
diff
command. Use
red
and a linewidth of 3.
c
. In the combined tiles
[2, 3, 5, 6]
, plot the phase plot with
?(?)
along the horizontal axis and its derivative
?′(?)
along the vertical axis. Use
black
and a linewidth of 3.
d.
You can see the phase plot approaches a
limiting ellipse
as the transitory parts of the solution die out for
large values of
t
. Add hat limiting ellipse now. It is drawn using just the particular solution:
?
𝑝
(?) = 4 ∙ cos(2?)
+
32 ∙ sin(2?)
and its derivative:
?
𝑝
′
(?) = −8?𝑖𝑛
+
64 ∙ cos(2?)
e.
Add
legend
as shown in the sample.
Adjust all the axes to get a good view of your plots using the
axis
command.
Questions: 9-10: Paste your completed multiplot in the answer template.
Replace
the image below with your completed multiplot. Some info has been obscured in the sample with a
dragon's eye! Your submission must not include the dragon eye.
Sample
Lab 5: Method of Undetermined Coefficients
ENGR-232 Winter 2022
8
Ready to Submit?
Be sure all ten questions are answered. When your lab is complete, be sure to submit three files:
1.
Your
completed Answer Template
as a PDF file
2.
A copy of your
MATLAB Live Script
3.
A
PDF
copy of your
MATLAB Live Script
(Save-
Export to PDF…)
The due date is the day after your lab section by
11:59pm
to receive full credit. You have one more day, to submit the lab (but
with a small penalty), and then the window closes for good and your grade will be zero.
Grader will award 2 points as follows.
i.
All three graphs are correct
ii.
The red limit cycle and the legend are included in the phase plot.