Math 2250
Summer 2010
Test 1
You have 50 minutes to complete this test. Calculators are not allowed. Be sure to show all work.
Each problem is worth 10 points.
1. Find the solution to the initial value problem
dy
= xe−x y(0) = 1.
dx
Since the right-hand side depends only on the independent variable x, we can solve by integrating
both sides to obtain
Z
y(x) = xe−x dx.
In order to compute the integral, we must use integration by parts. To this end, we make the
substitutions u = x and dv = e−x , so du = dx and v = −e−x . The integral then becomes
Z
−x
y(x) = −xe + e−x dx = −xe−x − e−x + C.
The initial condition implies that
1 = y(0) = −1 + C,
so C = 2. Therefore, the solution is
y(x) = −xe−x − e−x + 2 = e−x (1 − x) + 2.
2. Find the explicit general solution to the ODE
x
dy
− y = 2x2 y.
dx
The first thing that must be done is to decide which of the methods that we have studied will work
to solve this equation. Putting it in the form dy/dx = f (x, y), we have
2
2x + 1
dy
2x2 y + y
=
=y
.
dx
x
x
Since we have factored the right-hand side into a function of x only times a function of y only, this
is a separable equation. Separating the variables and integrating, we find that
Z
Z
dx2 + 1
dy
=
dx.
y
x
The integral on the left is
Z
dy
= ln |y|,
y
1
and the integral on the right is
Z
Z 2x2 + 1
1
dx =
2x +
dx = x2 + ln |x| + C.
x
x
Equating the two antiderivaties, we find that
ln |y| = x2 + ln |x| + C,
so
2
y(x) = Cex |x|.
3. A tank initially contains 100 gal of brine containing 50 lb of salt. Brine containing 1 lb of salt
per gallon enters the tank at a rate of 5 gal/sec, and the well mixed brine in the tank flows out at
a rate of 3 gal/s. Find a formula for the amount x of salt in the tank as a function of time.
First, recall that the change in the amount of salt in the tank with respect to time is equal to the
rate at which salt flows in minus the rate at which salt flows out. The rate at which salt flows in
is equal to the rate at which salt water flows in times the concentration of salt in the salt water.
Similarly, the rate at which salt flows out is equal to the rate at which salt water flows out times
the concentration of salt in the salt water. Therefore,
x
dx
= (1)(5) − 3
,
dt
2t + 100
where V (t) = (ri −r0 )t+V0 = 2t+100 is the volume of salt water in the tank at time t. Rearranging
the equation as
dx
3
+
x = 5,
dt
2t + 100
we see that this is a first order linear equation. The integrating factor for the equation is
Z
3
ρ(t) = exp
dt .
2t + 100
Making the substitution u = 2t + 100, this becomes
Z
3
3
3
du
3
ρ(t) = exp
ln |u| = u 2 = (2t + 100) 2 .
= exp
2
u
2
Multiplying through the equation by our integrating factor, it becomes
3
(2t + 100) 2
3
3
3
dx
+
(2t + 100) 2 x = 5(2t + 100) 2 .
dt
2t + 100
We now recognize the left-hand side as a derivative and write
h
i
3
3
Dt (2t + 100) 2 x = 5(2t + 100) 2 .
Integrating both sides with respect to t, we find
Z
3
3
(2t + 100) 2 x = 5(2t + 100) 2 dt.
2
Making the substitution u = 2t + 100, we have
Z
3
3
5
5
52 5
2
u 2 du =
u 2 + C = (2t + 100) 2 + C.
(2t + 100) x =
2
25
Dividing both sides of this equation by the integrating factor gives the solution
3
x(t) = (2t + 100) + C(2t + 100)− 2 .
We know that initially there are 50 lb of salt in the tank, so
3
50 = x(0) = 100 + C(100)− 2 ,
which implies that
C = −50000.
Therefore, the particular solution is
3
x(t) = (2t + 100) − 50000(2t + 100)− 2 .
4. Find all the critical points (equilibrium solutions) of the equation
dx
= x3 (x2 − 4)
dt
and classify each as either stable or unstable.
The equilibrium solutions are the zeros of the right-hand side function in this autonomous differential
equation. In other words, for the constant solution x0 to be an equilibrium solution, it must
satisfy
x30 (x20 − 4) = 0.
This implies that either x0 = 0 or x20 − 4 = 0, in which case x0 = ±2. Therefore, we have three
equilibrium solutions at −2, 0, and 2. In order to determine the stability of these critical points,
we look at the sign of the function f (x) = x3 (x2 − 4) between these points. A graph of f looks
like
Since f changes sign from negative to positive at −2, it is unstable. At 0, f goes from positive
to negative, so it is a stable fixed point. The change at 2 is from negative to positive, so 2 is an
unstable equilibrium.
3
5. Use Euler’s method to approximate the solution to the initial value problem
dy
= 2xy 2 y(0) = 1
dx
on the interval [0, 1/2] with step size h = 1/4.
The initial condition is y(0) = 1, so x0 = 0 and y0 = 1. Taking one step, we have that x1 = 1/4
and
1
y1 = y0 + hf (x0 , y0 ) = 1 +
2(0)(1)2 = 1.
4
We take another step to find x2 = 1/2 (so this is our last step) and
1
9
1
2
(1) = .
y2 = y1 + hf (x1 , y1 ) = 1 +
4
4
8
6. Find all solutions to the system of equations
2x1 − 2x2 − 6x3 = −4
4x1 − 3x2 − 10x3 = −5
2x1 − 2x3 = 2
We form the augmented matrix and row reduce:

 
..
2
−2
−6
.
−4

  1 −1 −3

 
..
 4 −3 −10 . −5  ∼  0 1
2

 
..
2 0
−2 . 2
0 2
4
 
..
. −2   1
 
..
∼
. 3 
  0
..
. 6
0
0
−1
1
2
0
0
..
.
..
.
..
.

1 

.
3 

0
Since there is not a leading one in the column for x3 , it is a free variable, so let us write x3 = r.
Substituting this expression, we find that
x3 = r, x2 = −2r + 3, x1 = r + 1.
7. Find the reduced row-echelon form of the matrix


2 12 4
 1 −2 1  .
3 0 2
We have that

2 12
 1 −2
3 0
8. Let
 
4
1
1 ∼ 0
2
0
A=
3
2
1
1
 
1 0
6
2
−8 −1  ∼  0 1
−18 −4
0 0
, B=
3
−1
1
−4
2
3
4
5
4
1
8
−4 +

18
8
, and C =

1
∼ 0
0
−1
−4
2
9
−4
3

0
0 .
1
0
1
0
.
Find A(2B + 3C).
We have that
A(2B + 3C) =
=
=
3
2
1
1
3
2
1
1
3
2
1
1
3
2
−1
6
−2
2
−8
3
−14
8
19
1
−4
+3
−1 2
−4 9
−4
3
−3 6 −12
−12 27
9
−8
−5 43 −9
=
.
15
−8 35 −1
4
6
2
3
+
9. Suppose that A is a m × n matrix and c1 and c2 are constant scalars. If ~x1 and ~x2 are both
solutions to the homogeneous system of equations with coefficient matrix A, show that c1 ~x1 + c2 ~x2
is also.
That ~x1 and ~x2 are solutions to the homogeneous system with coefficient matrix A means that
A~x1 = A~x2 = ~0. We must only check that c1 ~x1 + c2 ~x2 also satisfies the equation. To this end, we
have that
A(c1 ~x1 + c2 ~x2 ) = c1 A~x1 + c2 A~x2 = c1~0 + c2~0 = ~0.
Therefore c1 ~x1 + c2 ~x2 satisfies the homogeneous equation also.
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