Math 601, Solutions to Selected Problems
Solving Systems of Linear Differential Equations
The following problems come from Section 6.2 of the textbook Linear Algebra
and Vector Calculus at Texas A&M.
Note 1: For problem 1, your answer may be correct, but look different than
the answer given, because of the constants. Check to see if your eigenvectors
are multiples of the given eigenvectors.
Note 2: Since the other problems all have initial conditions, the answer
given is the only correct answer.
1
Problems
1. Find the general solution to each of the following systems.
(a)
y10 =
y1 + y2
y20 = −2y1 + 4y2
(e)
y10 = 3y1 − 2y2
y20 = 2y1 + 3y2
+ y3
y10 = y1
0
2y2 + 6y3
(f) y2 =
0
y3 =
y2 + 3y3
2. Solve each of the following initial value problems.
(a)
y10 = −y1 + 2y2
y20 = 2y1 − y2
y1 (0) = 3, y2 (0) = 1
(b)
y10 = y1 − 2y2
y20 = 2y1 + y2
y1 (0) = 1, y2 (0) = −2
− 6y3
y10 = 2y1
0
=
y
− 3y3
y
(c) 2
1
0
y2 − 2y3
y3 =
y1 (0) = y2 (0) = y3 (0) = 2
6. Solve the initial value problem
−2y2 + y10 + 2y20
y100 =
00
+ 2y10 − y20
y2 = 2y1
y1 (0) = 1,
, y10 (0) = −3,
y2 (0) = 0,
1
y20 (0) = 2
10. Three masses are connected by a series of springs between two fixed
points as shown in the figure. Assume that the springs all have the
same spring constant, and let x1 (t), x2 (t), and x3 (t) represent the displacement of the respective masses at time t.
(a) Derive a system of second-order differential equations that describes the motion of this system.
(b) Solve the system if m1 = m3 = 31 , m2 = 41 , k = 1, and
x1 (0) = x2 (0) = x3 (0) = 1
x01 (0) = x02 (0) = x03 (0) = 0
2
Solutions
1. Find the general solution to each of the following systems.
(a)
y10 =
y1 + y2
0
y2 = −2y1 + 4y2
(e)
y10 = 3y1 − 2y2
y20 = 2y1 + 3y2
y10 = y1
+ y3
2y2 + 6y3
(f) y20 =
y30 =
y2 + 3y3
Answer:
(a) This is a system of the form Y0 = AY where
µ
¶
1 1
A=
−2 4
The eigenvalues of this matrix
µ are
¶ λ1 = 2 andµ λ2 =
¶ 3 with corre1
1
sponding eigenvectors v1 =
and v2 =
.
1
2
µ ¶
µ ¶
1
1
2t
3t
Thus, e
and e
are both solutions to the system.
1
2
The general solution is all linear combinations of these two solutions.
µ
y1
y2
¶
µ
2t
= c1 e
1
1
¶
µ
+ c2 e
2
3t
1
2
¶
µ
=
c1 e2t + c2 e3t
c1 e2t + 2c2 e3t
¶
(e) This is a system of the form Y0 = AY where
µ
A=
3 −2
2
3
¶
The eigenvalues of this matrix are λ = 3 ± 2i.µ The
¶ eigenvector
i
corresponding to the eigenvalue λ = 3 + 2i is
. Using this
1
eigenvalue and eigenvector, we get the solution
µ
λt
e v = e
(3+2i)t
i
1
¶
µ
i
= e (cos(2t) + i sin(2t))
1
µ
¶
i cos(2t) − sin(2t)
3t
= e
cos(2t) + i sin(2t)
¶
3t
Separating into real and imaginary parts, we have
µ
µ ¶¶ µ
¶
i
−e3t sin(2t)
(3+2i)t
Re e
=
1
e3t cos(2t)
µ
µ ¶¶ µ 3t
¶
i
e cos(2t)
(3+2i)t
Im e
=
1
e3t sin(2t)
Any linear combination of these is a solution, so the general solution is
µ
y1
y2
¶
µ
−e3t sin(2t)
e3t cos(2t)
¶
µ
e3t cos(2t)
= c1
+ c2
e3t sin(2t)
µ
¶
−c1 e3t sin(2t) + c2 e3t cos(2t)
=
c1 e3t cos(2t) + c2 e3t sin(2t)
¶
(f) This is a system of the form Y0 = AY where


1 0 1
A= 0 2 6 
0 1 3
The eigenvalues of this matrix are 
λ1 = 0,
λ3 =1 with
 λ2 = 5, 
−1
1



−3 , v2 =
8 , and
corresponding eigenvectors v1 =
1
4
 
1
v3 =  0 .
0
3

 

 
 
−1
−1
1
1
Thus, e0t  −3  =  −3 , e5t  8 , and et  0  are all
1
1
4
0
solutions to the system. The general solution is all linear combinations of these three solutions.
 

 


1
y1
−1
1
 y2  = c1  −3  + c2 e5t  8  + c3 et  0 
0
4
y3
1


−c1 + c2 e5t + c3 et
=  −3c1 + 8c2 e5t 
c1 + 4c2 e5t

2. Solve each of the following initial value problems.
(a)
y10 = −y1 + 2y2
y20 = 2y1 − y2
y1 (0) = 3, y2 (0) = 1
(b)
y10 = y1 − 2y2
y20 = 2y1 + y2
y1 (0) = 1, y2 (0) = −2
y10 = 2y1
− 6y3
− 3y3
(c) y20 = y1
0
y3 =
y2 − 2y3
y1 (0) = y2 (0) = y3 (0) = 2
Answer:
(a) This is a system of the form Y0 = AY where
µ
A=
−1
2
2 −1
¶
The eigenvalues of this matrix are
λ2 =¶ 1 with
µ λ1 ¶= −3 and µ
1
1
.
corresponding eigenvectors v1 =
and v2 =
−1
1
µ
¶
µ ¶
1
1
−3t
t
Thus, e
and e
are both solutions to the sys−1
1
tem. The general solution is all linear combinations of these two
solutions.
µ
y1
y2
¶
µ
−3t
= c1 e
1
−1
¶
µ
+ c2 e
t
1
1
¶
The initial conditions tell us that
c1 + c2 = 3
−c1 + c2 = 1
4
µ
=
c1 e−3t + c2 et
−c1 e−3t + c2 et
¶
Solving, we get c1 = 1 and c2 = 2. Thus, the solution to the initial
value problem is
y1 = e−3t + 2et
y2 = −e−3t + 2et
(b) This is a system of the form Y0 = AY where
µ
A=
1 −2
2
1
¶
The eigenvalues of this matrix are λ = 1 ± 2i. The
µ eigenvector
¶
i
corresponding to the eigenvalue λ = 1 + 2i is v1 =
.
1
Using this eigenvalue and eigenvector, we get the solution
µ
λt
(1+2i)t
e v = e
i
1
¶
µ
i
= e (cos(2t) + i sin(2t))
1
µ
¶
i cos(2t) − sin(2t)
= et
cos(2t) + i sin(2t)
¶
t
Separating into real and imaginary parts, we have
µ
Re e
µ
i
1
(1+2i)t
µ
µ
(1+2i)t
Im e
i
1
¶¶
µ
=
¶¶
µ
=
−et sin(2t)
et cos(2t)
et cos(2t)
et sin(2t)
¶
¶
Any linear combination of these is a solution, so the general solution is
µ
y1
y2
¶
µ
−et sin(2t)
et cos(2t)
¶
µ
et cos(2t)
= c1
+ c2
et sin(2t)
µ
¶
−c1 et sin(2t) + c2 et cos(2t)
=
c1 et cos(2t) + c2 et sin(2t)
The initial conditions tell us that
c2 = 1
c1 = −2
5
¶
Thus, c1 = −2 and c2 = 1. Thus, the solution to the initial value
problem is
y1 = 2et sin(2t) + et cos(2t)
y2 = −2et cos(2t) + et sin(2t)
(c) This is a system of the form Y0 = AY where

2 0 −6
A =  1 0 −3 
0 1 −2

The eigenvalues of this matrix are λ1 
= 1, 
λ2 = 0, 
and λ
3 = −1
6
3
with corresponding eigenvectors v1 =  3 , v2 =  2 , v3 =
1
1
 
2
 1 .
1
 
   
 
6
3
3
2
t
0t 
−t 




3 ,e
2
2
1  are all soluThus, e
=
and e
1
1
1
1
tions to the system. The general solution is all linear combinations
of these three solutions.


 
 
 
y1
6
3
2
 y2  = c1 et  3  + c2  2  + c3 e−t  1 
y3
1
1
1


t
−t
6c1 e + 3c2 + 2c3 e

3c1 et + 2c2 + c3 e−t 
=
c1 et + c2 + c3 e−t
The initial conditions tell us that
6c1 + 3c2 + 2c3 = 2
3c1 + 2c2 + c3 = 2
c1 + c2 + c3 = 2
Solving, we get c1 = −1, c2 = 2, and c3 = 1. Thus, the solution
to the initial value problem is
6
y1 = −6et + 6 + 2e−t
y2 = −3et + 4 + e−t
y3 = −et + 2 + e−t
6. Solve the initial value problem
y100 =
−2y2 + y10 + 2y20
00
+ 2y10 − y20
y2 = 2y1
y1 (0) = 1,
, y10 (0) = −3,
y2 (0) = 0,
y20 (0) = 2
Answer: First, we set y3 = y10 and y4 = y20 , and we rewrite the equations as a system of linear differential equations:
y10 =
y3
0
y2 =
y4
0
y3 =
−2y2 + y3 + 2y4
y40 = 2y1
+ 2y3 − y4
This is a system of the form Y0 = AY where


0
0 1
0
 0
0 0
1 

A=
 0 −2 1
2 
2
0 2 −1
The eigenvalues of this matrix are λ1 = −2,λ2 =−1, λ3 
= 2, and

−1
−2
 1 
 1 



λ4 = 1 with corresponding eigenvectors v1 = 
 2 , v2 =  2 ,
−2
−1
 
 
1
1
 1 
 
, and v4 =  2 .
v3 = 
 2 
 1 
2
2
Thus, the general solution is






−1
−2
y1
 1 
 1 

 y2 
−2t 
−t 
2t 




 y3  = c1 e  2  +c2 e  2  +c3 e 
y4
−2
−1

The initial conditions tell us that
7


1

1 
 +c4 et 


2
2

1
2 

1 
2
−c1 − 2c2 + c3 + c4 = 1
c1 + c2 + c3 + 2c4 = 0
2c1 + 2c2 + 2c3 + c4 = −3
−2c1 − c2 + 2c3 + 2c4 = 2
Solving, we get that c1 = −1, c2 = 0, c3 = −1, and c4 = 1.
Thus, the solution to the initial value problem is
y1 = e−2t − e2t + et
y2 = −e−2t − e2t + 2et
10. Three masses are connected by a series of springs between two fixed
points as shown in the figure. Assume that the springs all have the
same spring constant, and let x1 (t), x2 (t), and x3 (t) represent the displacement of the respective masses at time t.
(a) Derive a system of second-order differential equations that describes the motion of this system.
(b) Solve the system if m1 = m3 = 31 , m2 = 41 , k = 1, and
x1 (0) = x2 (0) = x3 (0) = 1
x01 (0) = x02 (0) = x03 (0) = 0
Answer:
(a)
m1 x001 (t) = −kx1 + k(x2 − x1 )
m2 x002 (t) = −k(x2 − x1 ) + k(x3 − x2 )
m3 x003 (t) = −k(x3 − x2 ) − kx3
(b) This is a system of the form X00 = AX where


−6
3
0
4 
A =  4 −8
0
3 −6
8
The eigenvalues of this matrix are λ1 = −12, 
λ2 = 
−6, and
1
λ3 = −2, with corresponding eigenvectors v1 =  −2 , v2 =
1


 
1
3
 0 , and v3 =  4 .
−1
3
We will change to the basis {v1 , v2 , v3 }. Let the variables in the
new basis be called u1 , u2 , and u3 . Then, with respect to this
basis, we have the system of differential equations:
u001 = −12u1
u002 = −6u2
u003 = −2u3
Then, the solutions are:
√
√
u1 = c1 cos( 12t) + c2 sin( 12t)
√
√
u2 = c3 cos( 6t) + c4 sin( 6t)
√
√
u3 = c5 cos( 2t) + c6 sin( 2t)
Changing back to the original basis, the solutions are:




x1
1
³
´
√
√
 x2  = c1 cos( 12t) + c2 sin( 12t)  −2 
x3
1


1
³
√
√ ´
+ c3 cos( 6t) + c4 sin( 6t)  0 
−1
 
³
√
√ ´ 3
+ c5 cos( 2t) + c6 sin( 2t)  4 
3
Thus:
√
√
√








cos(2√3t)
sin(2√3t)
x1
cos( 6t)
 x2  = c1  −2 cos(2 3t)  + c2  −2 sin(2 3t)  + c3 

√ 0
√
√
x3
− cos( 6t)
cos(2 3t)
sin(2 3t)
√
√
√






3 cos(√2t)
3 sin(√2t)
sin( 6t)
 + c5  4 cos( 2t)  + c6  4 sin( 2t) 
+c4 
√ 0
√
√
− sin( 6t)
3 cos( 2t)
3 sin( 2t)
From the initial condition x1 (0) = x2 (0) = x3 (0) = 1, we get the
equations
9
c1 + c3 + 3c5 = 1
−2c1 +
4c5 = 1
c1 − c3 + 3c5 = 1
Solving, we get c1 = 0.1, c3 = 0, c5 = 0.3.
From the initial condition x01 (0) = x02 (0) = x03 (0) = 0, we get the
equations
√
√
√
2√3c2 + 6c4 + 3√2c6 = 0
−4√3c2 + √
4√2c6 = 0
2 3c2 − 6c4 + 3 2c6 = 0
Solving, we get c2 = 0, c4 = 0, and c6 = 0. Thus, the solution to
the initial value problem is
√
√

 

0.1 cos(2 √3t) + 0.9 cos( √2t)
x1
 x2  =  −0.2 cos(2 3t) + 1.2 cos( 2t) 
√
√
x3
0.1 cos(2 3t) + 0.9 cos( 2t)
10