MATH 2250—Final Exam
SOLUTIONS
Wednesday, December 17, 2008, 10:30AM–12:30PM
• Write your name and ID number at the top of this page.
• Show all your work.
• You may refer to one double-sided sheet of notes during the exam and
nothing else.
• Calculators are not allowed.
—Begin Final—
1. Solve the following initial value problem for the unknown function x(t).
dx
= x + x2 ,
dt
x(0) = 1
SOLUTION:
The equation is nonlinear, but it is separable:
Z
Z
dx
= dt
x + x2
The integral on the left-hand side can be done after decomposing the
integrand using partial fractions:
1
1
1
= −
x(x + 1)
x x+1
Then we integrate both sides and proceed:
Z
Z
dx Z dx
−
= dt
x
x+1
ln |x| − ln |x + 1| = t + c1
The expressions x and x + 1 are positive at the initial condition, so we
may remove the absolute value symbols and proceed:
x
= t + c1
ln
x+1
x
= et+c1 = c2 et , where c2 = ec1
x+1
Now’s a good time to plug in the initial condition:
1
= c2 e0 ,
1+1
so c2 = 1/2
x
1
= et
x+1
2
Finally, we solve for x to arrive at the solution:
x(t) =
1
1 t
e
2
− 12 et
2. At time t = 0 minutes, a tank contains 200 gallons of pure water.
Starting at t = 0, salt water is pumped into the tank at the rate of 2
gallons per minute, while the well-mixed solution flows out the other
end of the tank at the same rate. The concentration of salt in the
inflowing solution changes with time as follows:
2(1 − e−t/50 ) pounds of salt per gallon flowing in at time t
Find a formula for the amount (pounds) of salt in the tank at time t.
SOLUTION:
Let x(t) represent pounds of salt in the tank at time t. The rate of
change of salt in the tank is given by the “rate in” minus the “rate
out”:
Rate In: 2(1 − e−t/50 ) pounds/gallon × 2 gallons/minute
Rate Out: x(t) pounds/200 gallons × 2 gallons/minute
So our differential equation is
x
dx
= 4(1 − e−t/50 ) −
dt
100
The equation is not separable, but it is linear so we can use the integrating factor et/100 :
x t/100
dx t/100
e
+
e
= 4et/100 − 4e−t/100
dt
100
d t/100 xe
= 4et/100 − 4e−t/100
dt
xet/100 = 400et/100 + 400e−t/100 + C
x(t) = 400 + 400e−t/50 + Ce−t/100
The tank was filled with pure water initially, so our we apply the initial
condition x(0) = 0:
x(0) = 400 + 400 + C = 0 so C = −800
Then our solution is:
x(t) = 400(1 + e−t/50 − 2e−t/100 )
3. Given the matrix


1 0 −17
8 
A=
 0 1

1 2 −1
(a) Find A−1 , or show that it does not exist.
SOLUTION:
The inverse does not exist. One can prove this by showing that
the determinant of A is zero, or by showing that A is not rowequivalent to I, the 3 × 3 identity matrix:






1 0 −17
1 0 −17
1 0 −17





8 → 0 1 8 → 0 1 8 

 0 1
0 0 0
0 2 16
1 2 −1
Gaussian elimination produced a row of all zero’s, so A could not
be converted to the identity matrix.
(b) Express the first column of A as a linear combination of the other
two columns, or explain why it is not possible to do so.
SOLUTION:
We’re looking for a set of constants (c1 , c2 , c3 ) not all zero that
solves





0
c1
1 0 −17





8 
 0 1
  c2  =  0 
0
c3
1 2 −1
The Gaussian elimination above in part a) shows that c3 is a free
variable. If we choose c3 = 1, this results in c2 = −8 and c1 = 17.
This means that










1
0
−17
0



 



17  0  − 8  1  +  8  =  0 
1
2
−1
0
or




1
0
−17
1 
8  



 0 =
 1 −
 8 
17
17
2
−1
1
4. Find a general solution x(t) to the following differential equation.
x00 + x0 + x = 2t
SOLUTION:
First we find the complementary solution, or the solution to x00 +x0 +x =
0. Plugging in x = ert leads to the characteristic equation r2 +r+1 = 0.
Using the quadratic formula we get the roots
√
3
1
i
r=− ±
2
2
which corresponds to the complementary solution
√
√
xc = c1 e−t/2 cos( 3t/2) + c2 e−t/2 sin( 3t/2)
Next we need a particular solution that satisfies the full equation. The
right-hand side is a linear function, so the method of undetermined
coefficients is appropriate. There is no duplication with the complementary solution to worry about, so we plug in the form
xp = At + B
which leads to
A + At + B = 2t
so A = 2 and B = −2. The full solution is x = xc + xp , or
√
√
x(t) = c1 e−t/2 cos( 3t/2) + c2 e−t/2 sin( 3t/2) + 2t − 2
5. Find the function x(t) that is the inverse Laplace transform of each of
the following functions:
2
5
1
3
4
+ 2+ 3+
+ 2
s s
s
s+1 s +1
SOLUTION:
(a) X(s) =
X(s) =
1
1
1
32
1
+2 2 + 3 +4
+5 2
s
s
2s
s+1
s +1
3
x(t) = 1 + 2t + t2 + 4e−t + 5 sin t
2
s+9
(b) X(s) = 2
s + 6s + 13
SOLUTION:
The denominator is not factorable, so we complete the square:
X(s) =
=
s2
s+9
s+9
s+3
6
=
=
+
2
2
+ 6s + 9 + 4
(s + 3) + 4
(s + 3) + 4 (s + 3)2 + 4
6
s+3
2
s+3
+
=
+3
2
2
2
(s + 3) + 4 (s + 3) + 4
(s + 3) + 4
(s + 3)2 + 4
So
x(t) = e−3t cos 2t + 3e−3t sin 2t
6. Use the Laplace transform method to find the transformed solution
X(s) to the following initial value problem (you do not have to transform back to get the final solution).
x00 + 3x0 + x = 1 + f (t)
where
(
f (t) =
x0 (0) = 2,
x(0) = 1
0
, t<2
t−2 , t≥2
SOLUTION:
The function f (t) can be rewritten u(t − 2)g(t − 2), where g(t) = t.
Then, transforming both sides we get
s2 X − s − 2 + 3(sX − 1) + X =
1 e−2s
+ 2
s
s
1
e−2s
(s + 3s + 1)X = + s + 5 + 2
s
s
−2s 2
1/s + s + 5 + e /s
X(s) =
s2 + 3s + 1
2
7. Find a general solution (in terms of real functions) for x1 (t) and x2 (t)
that satisfies the following system of differential equations.
x02 = 2x1 − x2
x01 = 5x1 − 9x2
SOLUTION:
In vector–matrix form we have
"
0
x =
#
5 −9
2 −1
x
and we can use the eigenvalue method to find solutions of the form
x = veλt . Find eigenvalues λ first:
5−λ
−9
2
−1 − λ
→
→
=0
(5 − λ)(−1 − λ) + 18 = 0
λ2 − 4λ + 13 = 0
→
λ = 2 ± 3i
We choose one eigenvalue, λ = 2 − 3i, and find the corresponding
eigenvector v:
"
5 − (2 − 3i)
−9
2
−1 − (2 − 3i)
"
3 + 3i
−9
2
−3 + 3i
#
v=0
#
v=0
This is solved by v = (3 − 3i, 2) (among other vectors), which leads to
the following complex solution to the differential equation:
"
x=
3 − 3i
2
"
2t
=e
#
"
e
(2−3i)t
=
3 cos 3t − 3 sin 3t
2 cos 3t
#
3 − 3i
2
#
e2t (cos 3t − i sin 3t)
"
−3 cos 3t − 3 sin 3t
−2 sin 3t
+i
#!
The real part and the imaginary part of this solution form a basis for
the full solution space of the differential equation. So we may write the
general solution in real form as
"
x = c1 e
2t
3 cos 3t − 3 sin 3t
2 cos 3t
#
"
2t
+ c2 e
−3 cos 3t − 3 sin 3t
−2 sin 3t
#
or
x1 (t) = 3(c1 − c2 )e2t cos 3t − 3(c1 + c2 )e2t sin 3t
x2 (t) = 2c1 e2t cos 3t − 2c2 e2t sin 3t
8. For the following matrix,
"
A=
2 −4
1 −2
#
calculate the matrix exponential eAt in two different ways:
(a) By finding a fundamental matrix Φ(t) that solves X0 = AX and
using the formula eAt = Φ(t)Φ−1 (0)
SOLUTION:
We can construct a fundamental matrix by finding two solutions
to x0 = Ax using the eigenvalue method.
2−λ
−4
1
−2 − λ
→
=0
λ2 = 0
→
(2−λ)(−2−λ)+4 = 0
We have a repeated eigenvalue. Find an eigenvector v1 by solving
"
2 −4
1 −2
#
"
→
v1 = 0
v1 =
2
1
#
and then find a generalized eigenvector v2 by solving
"
2 −4
1 −2
#
"
v2 =
2
1
#
"
→
v2 =
#
"
1
0
#
and our two solutions are
"
2
1
#
"
0t
e
and
2
1
t+
1
0
#!
e0t
These two solutions make up the columns of Φ(t):
"
Φ(t) =
2 2t + 1
1
t
"
e
At
=
#
2 2t + 1
1
t
"
Φ(0) =
#"
2 1
1 0
#
#
"
0 1
1 −2
=
"
−1
Φ (0) =
0 1
1 −2
2t + 1 −4t
t
1 − 2t
#
#
(b) By showing that A is nilpotent and using the Taylor series formula
for eAt .
SOLUTION:
"
2
A =
2 −4
1 −2
#"
2 −4
1 −2
#
"
=
0 0
0 0
#
which means that A is nilpotent and the Taylor series formula for
eAt reduces to
"
eAt = I + At =
1 0
0 1
#
"
+
2 −4
1 −2
#
"
t=
1 + 2t −4t
t
1 − 2t
#