MATH 2243 Spring 2014 Sample Midterm Test I: Solutions

MATH 2243
Spring 2014
Sample Midterm Test I: Solutions
1. Find all solutions to the following differential equations.
a) y 2 y 0 = cos 6x, y(0) = 2.
Solution. First, we notice that this is a separable differential equation. Indeed,
we can rewrite it as
y 2 dy = cos 6x dx.
Taking integrals on both sides, we obtain
Z
Z
2
y dy = cos 6x dx
1
y3
= sin 6x + C.
3
6
Hence, y =
q
3
sin 6x
2
+ C.
The value of C can be found from the initial condition:
r
√
3 sin(6 · 0)
3
y(0) = 2 ⇒
+ C = C = 2.
2
Thus C = 8 and the solution is
r
3
y=
b) y 0 = x + e−x
2 /2
sin 6x
+ 8.
2
− xy.
Solution. We recognize that this is a first-order linear differential equation. Indeed, we can rewrite the given equation as
2
y 0 + |{z}
x ·y = x
e−x /2} .
| + {z
P (x)
(1)
Q(x)
Now, we can use our standard algorithm for solving this type of differential equations.
First, we find the integrating factor:
R
I(x) = e
P (x) dx
=e
R
x dx
= ex
2 /2
.
Next, we multiply both sides of (1) by I(x). As we’ve learned in class, the left-hand
side then can be written as [I(x) · y]0 . More concretely, in our case:
[ex
2 /2
· y]0 = xex
1
2 /2
+1
Finally, we integrate both sides
Z
2
x2 /2
e
· y = (xex /2 + 1) dx =
2
x /2
e|{z}
+x + C
use a u-sub to get this
and divide by I(x) = ex
2 /2
:
y =1+
x+C
← the general solution.
ex2 /2
2. The radioactive isotope Indium-111 is often used for diagnosis and imaging in nuclear
medicine. Its half life is 2.8 days. What was the initial mass of the isotope before
decay, if the mass in 2 weeks was 6g?
Solution. Let’s recall first that the differential equation describing the radioactive
decay process is
y 0 = ky,
where y = y(t) is the mass of the radioactive sample and k < 0 is a constant depending
on the half-life. More precisely, k = − lnτ 2 , where τ is the length of the half-life period
(if this is not totally obvious, work it out).
As we’ve seen in class before, the general solution of this separable differential equation
is
y = Cekt ,
where the constant C actually represents the initial amount of material y(0).
In our case, the statement that the mass of the sample became 6g after 2 weeks means
that (assuming that t is measured in days)
k
z }| {
ln 2
y(14)
−
·14
z}|{
2.8
6 = Ce
.
Solving for C = y(0), we get
C=
6
e− ln 2·5
= 6 · 32 = 192g.
3. A commercial fishery is estimated to have carrying capacity of 10 thousand pounds of
certain kind of fish. Suppose the annual growth rate of the total fish population P ,
measured in thousand pounds, is governed by the logistic equation
dP
P
= 1−
P
dt
10
and initially there is a total of 2 thousand lbs of fish. What is the fish population after
1 year?
Solution. The logistic equation is separable. Thus, we start by rewriting the given
equation as
dP
= dt
P
1 − 10
P
or simply
10
dP = dt
(10 − P ) P
Now, we would like to take the integral on both sides. While the right-hand side is
trivial, on the left-hand side we will need to use a partial fraction decomposition (that’s
a common technique for integrating rational functions).
So we write down the desired decomposition:
10
A
B
=
+ .
(10 − P ) P
10 − P
P
(2)
Our goal is to figure out for which values of A and B this all works. To this end, we
multiply both sides of (2) by (10 − P ) P and simplify:
10 = A · P + B · (10 − P ).
Setting P = 0, we see that B = 1, and if we take P = 10, we get A = 1. Hence,
Z
Z
1
1
10
dP =
+ dP = − ln |10 − P | + ln |P | + a constant.
(10 − P ) P
10 − P
P
The integral on the right-hand side is
Z
dt = t + C
and, combining, we get
− ln |10 − P | + ln |P | = t + C.
(3)
How to get rid of the absolute values in this expression? It is given that P (0) = 2 (we
start with 2,000 lbs of fish). Hence, 10−P (0) > 0 and P (0) > 0. Then |10−P | = 10−P
and |P | = P . So, we can rewrite (3) as
ln(P ) − ln(10 − P ) = t + C ⇒ ln
P
= t + C.
10 − P
Taking the exponent of both sides, we get
P
= et+C = et · eC = et · D.
10 − P
(4)
To find the value of D, we use the initial condition P (0) = 2:
P (0)
2
1
= = e0 · D ⇒ D = .
10 − P (0)
8
4
Then, returning to (4), we get
et
et
t
P = (10 − P ) ⇒ P = 2.5e − P ⇒ P
4
4
Finally, we find
P (t) =
2.5et
t
1 + e4
et
1+
4
= 2.5et .
4. Write the augmented coefficient matrix of the following linear system and transform it
to echelon form:


2x1 + 4x2 − x3 − 2x4 + 2x5 = 6
x1 + 3x2 + 2x3 − 7x4 + 3x5 = 9


5x1 + 8x2 − 7x3 + 6x4 + x5 = 4
Use the echelon matrix in (1) to solve the system by back substitution.
Solution. The augmented coefficient matrix is


2 4 −1 −2 2 6
 1 3 2 −7 3 9  .
5 8 −7 6 1 4
It can be transformed into an echelon form as follows1 :




1 3 2 −7 3 9
2 4 −1 −2 2 6
r1 ,r2
−2r1 +r2
 1 3 2 −7 3 9  swap
−→  2 4 −1 −2 2 6  −→
5 8 −7 6 1 4
5 8 −7 6 1 4




1 3
2 −7 3
9
1 3
2 −7 3
9
1 +r3
 0 −2 −5 12 −4 −12  −5r
 0 −2 −5 12 −4 −12  −4·r
−→
−→2
5 8 −7 6
1
4
0 −7 −17 41 −14 −41




9
9
1 3
2
−7
3
1 3
2 −7 3
r3 +r2
7r2 +r3
 0 8
 0 1
20 −48 16
48  −→
3 −7 2
7  −→
0 −7 −17 41 −14 −41
0 −7 −17 41 −14 −41




1 3 2 −7 3 9
1
3
2
−7
3
9
1
·r
4 3
 0 1 3 −7 2 7  −→
 0 1 3 −7 2 7 
0 0 4 −8 0 8
0 0 1 −2 0 2
From the echelon form we see that we can take x4 and x5 as free variables. So, let
x4 = s, x5 = t. Then, starting from the last line of the echelon form, we obtain
x3 = 2 + 2x4 − 0 · x5 = 2 + 2s
x2 = 7 − 3x3 + 7 · x4 − 2x5 = · · · = 1 + s − 2t
x1 = 9 − 3x2 − 2x3 + 7x4 − 3x5 = · · · = 2 + 3t
So the general solution of the system is
x1 = 2 + 3t, x2 = 1 + s − 2t, x3 = 2 + 2s, x4 = s, x5 = t,
where s, t are parameters.
1