Math 53 — Second Midterm
Page 1
Math 53, Fall 2007 — Second Midterm
Solutions
1. (a) Solve the following differential equation :
y 00 − 2y 0 + 2y = 2 et cos(t)
(b) Solve the corresponding initial value problem with initial condition :
y(0) = 0 ,
y 0 (0) = 1.
(a) One possible method is by using undetermined coefficients.
Step 1 : find solution to the homogeneous equation y 00 − 2y 0 + 2y = 0.
The characteristic equation is λ2 − 2λ + 2 = 0, which gives λ = 1 ± i. Thus, the general
solution to the homogeneous equation is :
yh = A et cos(t) + B et sin(t).
Step 2 : guess at the form of a solution. We note that the natural guess is a linear combination
of et cos(t) and of et sin(t), but these are in the solution space to the homogeneous equation.
So, we try for a particular solution of the form yp = At et cos(t) + Bt et sin(t). Then we have :
yp00 − 2yp0 + 2yp = −2A et sin(t) + 2B et cos(t).
Thus, we must take A = 0 and B = 1.
Step 3 : the general solution is then given by
A et cos(t) + B et sin(t) + t et sin(t).
(b) We must now solve for A and B. First, we compute :
y 0 (t) = et (A cos(t) + B sin(t) + t sin(t) − A sin(t) + B cos(t) + sin(t) + t cos(t))
Thus, y 0 (0) = A + B. Furthermore, y(0) = A. Thus, A = 0 and B = 1. The solution is then
given by :
y(t) = et (sin(t) + t sin(t).
(Alternate methods included using the method of variation of parameters.)
Math 53 — Second Midterm
Page 2
2. Solve the initial value problem :

 x0 = x − 4y
 y0 = x + y

 x(0) = −1
with
.
 y(0) = 1
Let
1 −4
A=
1
!
1
.
The characteristic polynomial is :
p(λ) = λ2 − 2λ + 5.
The eigenvalues are then λ± = 1 ± 2i. We now solve for eigenvectors, and obtain :
!
±2i
v± =
.
1
From this, we obtain the two complex valued fundamental solutions :
z+ (t) = eλ+ t v+
and
z− (t) = eλ− t v− .
From here, we have at least two different ways of solving the initial value problem.
Method 1 : finding real solutions, and working with these.
The real and imaginary parts of z+ represent solutions to the differential equation.
!
2i
z+ = et (cos(2t) + i sin(2t))
1
!
!!
−2 sin(2t)
2 cos(2t)
t
=e
+i
.
cos(2t)
sin(2t)
Thus, we obtain two real valued solutions to the differential equation :
!
!
−2
sin(2t)
2
cos(2t)
u(t) = et
and v(t) = et
.
cos(2t)
sin(2t)
These are linearly independent, and thus we need to find constants so that Au + Bv solves
the initial value problem. This gives us :
!
−1
1
=
2B
A
!
.
Thus, B = −1/2 and A = 1. The solution to the initial value problem is thus :
!
−2
sin(2t)
−
cos(2t)
et
cos(2t) − 21 sin(2t)
Math 53 — Second Midterm
Page 3
Method 2 : working with the complex solutions.
The desired solution is z(t) = Az+ (t)+Bz− (t) where B = Ā and A, B are complex coefficients.
We need z(0) = (−1, 1). Thus, Av+ + Āv− = (−1, 1). Put A = a + ib. Then, −1 = −4b and
1 = 2a. Thus, a = 1/2 and b = 1/4. The solution is then :
1
1
1
1
z(t) =
1 + i z+ +
1 − i z−
2
2
2
2
1
= Re( 1 + i z+ )
2
1
= Re(z+ ) − Im(z+ )
2
!
!!
−2
sin(2t)
2
cos(2t)
1
= et
−
2 sin(2t)
cos(2t)
!
t −2 sin(2t) − cos(2t)
=e
cos(2t) − 21 sin(2t)
.
3. (a) Let A be the matrix A =
(b) Let B be the matrix B =
3 2
0 3
1
!
. Find the matrix exponential eAt .
2
−2 5
!
. Find the matrix exponential eBt .
(a) Method 1 : Observe that 3 is the only eigenvalue of A. Notice that (A − 3 Id)2 = 0. Then
:
eAt = e3 Id t+(A−3 Id)t
= e3t
= e3t
Id +
0 2
0 0
!
1 2t
.
0 1
! !
t
Method 2 : Solve the initial value problem y0 = Ay and y(0) = (a, b) : y20 = 3y2 so y2 (t) =
e3t b. Then, y10 = 3y1 + 2y2 = 3y1 + 2 e3t b. By using the integrating factor e−3t , we obtain
y1 e−3t = 2bt + C. Thus, y(t) = ((a + 2bt) e3t , b e3t ). Writing this as a matrix :
! !
e3t 2t e3t
a
.
y(t) =
3t
0
e
b
Math 53 — Second Midterm
Page 4
(b) Method 1 : The characteristic polynomial is λ2 − 6λ + 9, and thus B has 3 as its only
eigenvalue. Then, we calculate that (B − 3 Id)2 = 0. Thus,
eBt = e3 Id t+(B−3 Id)t
= e3t (Id +(B − 3 Id)t)
! !
−2 2
3t
=e
Id +
t
−2 2
!
1
−
2t
2t
= e3t
−2t 1 + 2t
Method 2 : Change to a basis of generalized eigenvectors.
We find that v1 = (1, 1) is an eigenvector of B, and that v2 = (0, 1/2) is a generalized
eigenvector with
(B − 3 Id)v2 = v1 .
Let
S
−1
1 0
=
!
1
2
1
Then, we obtain :
1
S=
0
!
.
−2 2
Then,
SBS
−1
3 1
=
!
.
0 3
Thus,
0
1
B3 1 C
@
e
Bt
=S
−1
e
= S −1 e3t
3t
=e
0 3
At
1 t
0 1
S
!
(1)
S
1 − 2t
2t
−2t
2t + 1
(2)
!
.
(3)
(Note : a simpler solution would have been to notice that A and B are similar and reduce to
case (a). To do this, choose (0, 1) as the generalized eigenvector to put in our new basis.)
Math 53 — Second Midterm
Page 5
4. Solve the initial value problem :
 0
t

 x = x − y + z + e cos(t)

y 0 = y + 2z − et



z 0 = z.
subject to the initial condition x(0) = −1, y(0) = 1 and z(0) = 3.
Method 1 : brute force. We note this problem is upper triangular, so we may solve it by
starting at the last equation and working upwards.
z 0 = z has solution z(t) = C et . To satisfy the initial condition, we need z(t) = 3 et .
Then, y 0 = y + 6 et − et = y + 5 et . Using the integrating factor e−t we obtain : e−t y = 5t + C.
With the initial condition, we obtain y(t) = (5t + 1) et .
Finally, x0 = x − (5t + 1) et +3 et + et cos(t). Using the integrating factor e−t , we obtain :
R
e−t x(t) = −5t + 2 + cos(t)dt = − 25 t2 + 2t + sin(t) + C. With the initial condition, we obtain
x(t) = (− 25 t2 + 2t + sin(t) − 1) et .
The solution is then :



et (−1 + 2t − 52 t2 + sin(t)

 

t (5t + 1)
y(t) = 
.
e

 

z(t)
3 et
x(t)

Method 2 : variation of parameters using the matrix exponential.
Consider the matrix


1 −1 1



A=
0
1
2


0 0 1
We compute
eAt


1 −t −t2 + t


.
= et 
0
1
2t


0 0
1
We now look for a solution to the differential equation :
y0 = Ay + f (t)
where f (t) = (et cos(t), − et , 0). We will look for a solution of the form y(t) = eAt v(t). We
then have :
v0 (t) = e−At f (t)


cos(t) − t



=
 −1  .
0
Math 53 — Second Midterm
Page 6
Thus,

sin(t) −

v(t) = 

t2
2
+C
−t + C 0
C 00


.

The general solution to the differential equation is then :



2
1 −t −t2 + t
sin(t) − t2 + C




.
y(t) = et 
2t 
−t + C 0
0 1


0
0
C 00
1
To satisfy the initial condition, we must take C = −1, C 0 = 1 and C 00 = 3. This gives us :



t2
2
1 −t −t + t
sin(t) − 2 − 1



t



v(t) = e 0 1
−t + 1
2t  

3
0 0
1


sin(t) − 52 t2 + 2t − 1


.
= et 
5t
+
1


3
5. Consider the following system of two masses and three springs, sliding on a frictionless tabletop.
k1 = 2
k2 = 1
k3 = 2
m2
m1
Assume each mass is of 1 kg, and the spring constants are as indicated. Then, the positions
of the two masses are governed by the following system of differential equations :

 x00 = −3x + y
 y 00 = −3y + x
where x and y denote the positions of the two masses relative to the equilibrium point of this
system (which occurs at x = y = 0; we take the positive direction to be to the right).
(a) Transform this system into a first order system of differential equations.
(b) Find the solution with initial condition x(0) = 0, x0 (0) = 1, y(0) = 0 and y 0 (0) = 1.
Math 53 — Second Midterm
For this

0

−3

0

1
Page 7
problem, you may use without proof that :
 
 

1 0 0
i
i
0
 
 

 
 

0 1 0
 −2 = 2i −2 and −3
 
 −i 
0
0 0 1
  −i 
 

0 −3 0
2
2
1

 
−i
−i
√ 
 √ 
√  
 
0 1 0
  2 = i 2  2
 −i 
 
0 0 1
 
  −i 
√
√
2
2
0 −3 0
1
0
0

(a) Let p = x0 and q = y 0 . Then, we have the system :
x0 = p
p0 = −3x + y
y0 = q
q 0 = −3y + x.
Equivalently, this is :
 0 
x
0
 

p

  = −3
y 
0
 

q
1
 
x
 
 
0 1 0
 p .


0 0 1 
y 
0 −3 0
q
1
0
0
√
(b) From the given information, we have that ±2i and ± 2i are eigenvalues of this matrix.
Let

i

 
−2

v1 = 
 −i 
 
2

and

−i
√ 
 2

v2 = 
 −i 
 
√
2
We then have that each of the real part and imaginary part of each of e2it v1 and e
√
2it v
2
give four linearly independent solutions to the differential equation.
This gives :

0


1


0


1

 
 
 
 
−2
0
−2
0







x1 = cos(2t)   − sin(2t)   and y1 = sin(2t)   + cos(2t) 
−1
0
 
0
−1
0
0
2
2
 
 
 
 
0
−1
0
−1
√ 
 
√ 
 
√  2
√ 0
√  2
√ 0

 
 
 
x2 = cos(t 2) 
 0  − sin(t 2) −1 and y2 = sin(t 2)  0  + cos(t 2) −1 .
 
 
 
 
√
√
2
0
2
0
Math 53 — Second Midterm
Page 8
We want to find a solution
u(t) = Ax1 + By1 + Cx2 + Dy2
so that u(0) = (1, 0, 1, 0). This gives the system of equations :
B−D =0
√
−2A + C 2 = 1
−B − D = 0
√
2A + C 2 = 1
Thus, B = D = 0 and A = 0, C =
Thus,
√
2/2.

√
sin(t 2)
√ 
√ √
2 cos(t 2)
2


u(t) =
√

2 
sin(t
2)


√
√
2 cos(t 2)

The solution is then given by :
√
√
2
sin(t 2)
√2
√
2
y(t) =
sin(t 2).
2
x(t) =