Today
• The geometry of homogeneous and nonhomogeneous matrix equations
• Solving nonhomogeneous equations
• Method of undetermined coefficients
1
Monday, January 26, 2015
Second order, linear, constant coeff, nonhomogeneous (3.5)
• Our next goal is to figure out how to find solutions to nonhomogeneous
equations like this one:
y �� − 6y � + 8y = sin(2t)
• But first, a bit more on the connections between matrix algebra and
differential equations . . .
2
Monday, January 26, 2015
Some connections to linear (matrix) algebra
• An mxn matrix is a gizmo that takes an n-vector and returns an mvector:
y = Ax
• It is called a linear operator because it has the following properties:
A(cx) = cAx
A(x + y) = Ax + Ay
• Not all operators work on vectors. Derivative operators take a function
and return a new function. For example,
d2 y
dy
z = L[y] = 2 − 2
+y
dt
dt
• This one is linear because
L[cy] = cL[y]
L[y + z] = L[y] + L[z]
Monday, January 26, 2015
Note: y, z are functions
of t and c is a constant.
3
Some connections to linear (matrix) algebra
• A homogeneous matrix equation has the form
Ax = 0
• A non-homogeneous matrix equation has the form
Ax = b
• A homogeneous differential equation has the form
L[y] = 0
• A non-homogeneous differential equation has the form
L[y] = g(t)
4
Monday, January 26, 2015
Solutions to homogeneous matrix equations
• The matrix equation
Ax = 0
could have (depending on A)
(A) no solutions.
(B) exactly one solution.
(C) a one-parameter family of solutions.



1x = 0 1 0


−1
x
=
C



x = C1 −1 + C2 2
1 1
1
(D) an n-parameter family of solutions.
Choose the answer that is incorrect.
Possibilities:


5
Monday, January 26, 2015
Solutions to homogeneous matrix equations
Ax = 0 where


1 2
3
A = 1 −1 −2
2 1
1
• Example 1. Solve the equation
• Row reduction gives
• so
1
x1 − x3 = 0
3

1 0
A ∼ 0 1
0 0
and

−1/3
5/3 
0
5
x2 + x3 = 0
3
and
Each equation
describes a plane.
In this case, only
two of them really
matter.
x3 can be whatever
(because it doesn’t have a leading one).
6
Monday, January 26, 2015
Solutions to homogeneous matrix equations
• Example 1. Solve the equation
• so
1
x1 − x3 = 0
3
and
1
x1 = x3
3
5
x2 = − x3
3
Ax = 0 .
5
x2 + x3 = 0
3
x3 can be whatever.
1
x1 = C
3
5
x2 = − C
3
x3 = C
• Thus, the solution can be written as
Monday, January 26, 2015
and


1
C�  
x = C −5 .
3
3
7
Solutions to homogeneous matrix equations
Ax
=
0
where


1 −2 1
A =  2 −4 2 
−1 2 −1


1 −2 1
A ∼ 0 0 0
0 0 0
• Example 2. Solve the equation
• Row reduction gives
• so
x1 − 2x2 + x3 = 0
Monday, January 26, 2015
and both
x2 and x3 can be whatever.
 
 
2
−1
x = C1 1 + C2  0 
0
1
8
Solutions to homogeneous matrix equations
Ax
=
0
where


1 −2 1
A =  2 −4 2 
−1 2 −1


1 −2 1
A ∼ 0 0 0
0 0 0
• Example 2. Solve the equation
• Row reduction gives
• so
x1 − 2x2 + x3 = 0
Monday, January 26, 2015
x2 and x3 can be whatever.
 
 
2
−1
x = C1 1 + C2  0 
0
1
and both
9
Solutions to non-homogeneous matrix equations
• Example 3. Solve the equation

1
A = 1
2
• Row reduction gives
• so
2
−1
1

Ax = b

3
−2
1
1 0
0 1
0 0
and
where
 
2
b = 0 .
2
�

−1/3�� 2/3
5/3 �� 2/3
0 � 0
1
2
5
2
x1 − x3 = and x2 + x3 =
3
3
3
3
and
x3 can be whatever.
10
Monday, January 26, 2015
Solutions to non-homogeneous matrix equations
Ax = b .
5
2
x2 + x3 = and x3 can be whatever.
3
3
5
2
x2 = − x3 +
3
3
• Example 3. Solve the equation
1
• so x1 − x3 =
3
1
x1 = x3 +
3

2
and
3
2
3

 
2/3
1
C�    
x = C −5 + 2/3
3
0
3
the general solution to
the homogeneous
problem
Monday, January 26, 2015
one particular solution
to nonhomogeneous
problem
11
Solutions to nonhomogeneous differential equations
• To solve a nonhomogeneous differential equation:
1. Find the general solution to the associated homogeneous
problem, yh(t).
first order DE
second order DE
2. Find a particular solution to the nonhomogeneous problem, yp(t).
3. The general solution to the nonhomogeneous problem is their
sum:
y = yh + yp = C1 y1 + C2 y2 + yp
• For step 2, try “Method of undetermined coefficients”...
12
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 4. Define the operator
general solution to
L[y] = e
2t
L[y] = y + 2y − 3y.
��
. That is,
�
Find the
y �� + 2y � − 3y = e2t .
• Step 1: Solve the associated homogeneous equation
y �� + 2y � − 3y = 0.
yh (t) = C1 e + C2 e
−3t
t
2t
· ] to get e
• Step 2: What do you have to plug in to L[
• Try
yp (t) = Ae .
• L[yp (t)]
2t
= L[Ae2t ] =



(A)
5e
(B)
5Ae
2t
2t
out?
(C)
4e
(D)
4Ae
2t
2t
• A is an undetermined coefficient (until you determine it).
Monday, January 26, 2015
13
Method of undetermined coefficients (3.5)
• Example 4. Define the operator
general solution to
L[y] = y + 2y − 3y.
��
L[y] = e
2t
. That is,
• Summarizing:
�
Find the
y �� + 2y � − 3y = e2t .
• We know that, for any C1 and C2,
L[C1 et + C2 e−3t ] = 0
• We also know that
L[Ae2t ] = 5Ae2t
• Finally, by linearity, we know that
L[C1 e + C2 e
t
−3t
+ Ae ] = 0 + 5Ae
2t
2t
• So what’s left to do to find our general solution? Pick A =
=?1/5.
14
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 5. Find the general solution to the equation
y �� − 4y = et.
• What is the solution to the associated homogeneous equation?
(A)
yh (t) = C1 e + C2 e
(B)
yh (t) = C1 cos(2t) + C2 sin(2t)
2t
−2t
(C) yh (t)
= C1 e + C2 te
(D) yh (t)
= C1 cos(2t) + C2 sin(2t) + e
2t
2t
t
(E) Don’t know.
15
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 5. Find the general solution to the equation
y �� − 4y = et.
• What is the form of the particular solution?
(A)
yp (t) = Ae
(B)
yp (t) = Ae
(C)
yp (t) = Ae
(D)
yp (t) = Ate
2t
−2t
t
t
(E) Don’t know
16
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 5. Find the general solution to the equation
y �� − 4y = et.
• What is the value of A that gives the particular solution (Ae
t
)?
(A) A = 1
(B) A = 3
(C) A = 1/3
(D) A = -1/3
(E) Don’t know.
17
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 7. Find the general solution to the equation y ��
− 4y = e2t.
• What is the solution to the associated homogeneous equation?
(A)
yh (t) = C1 e + C2 e
(B)
yh (t) = C1 cos(2t) + C2 sin(2t)
2t
−2t
m
a
ex
2t
t
s
C
te
(C) yh (t) = C1 e + a
2
l
e
h
t
2t
−2t
t
s
a
= C1 e + C2 e
+e
(D) yh (t)
e
m
a
S
2t
e
l
p
(E) Don’t know.
18
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 7. Find the general solution to the equation y ��
− 4y = e2t.
• What is the form of the particular solution?
(A)
yp (t) = Ae
(B)
yp (t) = Ae
(C)
yp (t) = Ate
2t
−2t
2t
(Ae2t )�� − 4Ae2t = 0 !
• Simpler example in which
the RHS is a solution to the
homogeneous problem.
y −y =e
�
(D)
yp (t) = Ae
(E)
yp (t) = Ate
t
t
t
e−t y � − e−t y = 1
y = tet + Cet
• General rule: when your guess at yp makes LHS=0, try multiplying it by t.19
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 7. Find the general solution to the equation y ��
− 4y = e2t.
• What is the value of A that gives the particular solution (Ate
2t
(A) A = 1
(B) A = 4
(C) A = -4
(D) A = 1/4
(E) A = -1/4
)?
(Ate ) = Ae + 2Ate
2t �
2t
2t
(Ate2t )�� = 2Ae2t + 2Ae2t + 4Ate2t
= 4Ae + 4Ate
�
�
�
�
��
2t
2t
2t
Ate
− 4 Ate = 4Ae
2t
Need:
2t
=e
2t
20
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 8. Find the general solution to
y − 4y = cos(2t) .
��
• What is the form of the particular solution?
(A)
yp (t) = A cos(2t)
(B)
yp (t) = A sin(2t)
(C)
yp (t) = A cos(2t) + B sin(2t)
(D)
yp (t) = t(A cos(2t) + B sin(2t))
(E)
yp (t) = e (A cos(2t) + B sin(2t))
2t
Challenge: What small change to the DE makes (D) correct?
Monday, January 26, 2015
21
Method of undetermined coefficients (3.5)
• Example 8. Find the general solution to
y + y − 4y = cos(2t).
��
�
• What is the form of the particular solution?
(A)
yp (t) = A cos(2t)
(B)
yp (t) = A sin(2t)
(C)
yp (t) = A cos(2t) + B sin(2t)
(D)
yp (t) = t(A cos(2t) + B sin(2t))
(E)
yp (t) = e (A cos(2t) + B sin(2t))
2t
22
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 9. Find the general solution to
y �� − 4y = t3 .
• What is the form of the particular solution?
(A)
yp (t) = At
(B)
yp (t) = At + Bt + Ct
(C)
yp (t) = At3 + Bt2 + Ct + D
(D)
yp (t) = At3 + Be2t + Ce−2t
3
3
(E) Don’t know.
2
waste of time including
solution to homogeneous eq.
23
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• When RHS is sum of terms:
y �� − 4y = cos(2t) + t3
yp (t) = A cos(2t) + B sin(2t) + Ct3 + Dt2 + Et + F
24
Monday, January 26, 2015
Method of undetermined coefficients (3.5)
• Example 10. Find the general solution to
y + 2y = e + t
��
�
2t
3
.
• What is the form of the particular solution?
(A)
yp (t) = Ae2t + Bt3 + Ct2 + Dt
(B)
yp (t) = Ae2t + Bt3 + Ct2 + Dt + E
(C)
(D)
yp (t) = Ae2t + (Bt4 + Ct3 + Dt2 + Et)
yp (t) = Ae2t + t(Bt3 + Ct2 + Dt + E)
yp (t) = Ae2t + Be−2t + Ct3 + Dt2 + Et + F
(E)
yp (t) = Ae + Bte + Ct + Dt + Et + F
2t
2t
3
2
For each wrong answer, for what DE is it the correct form?
Monday, January 26, 2015
25
Method of undetermined coefficients (3.5)
• Example 11. Find the general solution to
y − 4y = t e
��
3 2t
.
• What is the form of the particular solution?
Monday, January 26, 2015
(A)
yp (t) = (At + Bt + Ct + D)e
(B)
yp (t) = (At3 + Bt2 + Ct)e2t
(C)
yp (t) = (At3 + Bt2 + Ct)e2t
3
2
−2t
+(Dt + Et + F t)e
(D)
yp (t) = (At4 + Bt3 + Ct2 + Dt)e2t
3
2
2t
yp (t) = t(At + Bt + Ct + D)e
(E)
yp (t) = (At4 + Bt3 + Ct2 + Dt + E)e2t
3
2
2t
26
Method of undetermined coefficients (3.5)
• Summary - finding a particular solution to L[y] = g(t).
• Include all functions that are part of the g(t) family (e.g. cos and sin)
• If part of the g(t) family is a solution to the homogeneous (h-)problem,
use t x (g(t) family).
• If t x (part of the g(t) family), is a solution to the h-problem, use t2 x (g
(t) family).
• For sums, group terms into families and include a term for each.
• For products of families, use the above rules and multiply them.
• If your guess includes a solution to the h-problem, you may as well
remove it as it won’t survive L[ ] so you won’t be able to determine its
undetermined coefficient.
27
Monday, January 26, 2015