Math 2250–016 (Underdown)
Exam 1
September 25, 2014
Name
1. [10 pts.] Solve the following nonlinear IVP explicitly for y(x)
dy
x−5
=
dx
y2
y(0) = 2.
Solution: This equation is separable.
Z
Z
2
y dy = (x − 5) dx
x2
y3
=
− 5x + C
3
2
2
1/3
3x
y(x) =
− 15x + C
2
2 = (0 − 0 + C)1/3
C=8
2
1/3
3x
y(x) =
− 15x + 8
2
Math 2250–016 (Underdown)
Exam 1, Page 2 of 8
September 25, 2014
2. [10 pts.] Find the general solution to
2
1 dy
− 2 y = x cos x,
x dx x
x > 0.
Solution: This equation is linear, so we’ll use the integrating factor method.
dy
2
− y = x2 cos x
dx x
R
R 2
I(x) = e p(x) dx = e − x dx = e−2ln|x| = x−2
x−2
dy
− 2x−3 y = cos x
dx
d −2
(x y) = cos x
dx
x−2 y = sin x + C
y = x2 sin x + Cx2
Math 2250–016 (Underdown)
Exam 1, Page 3 of 8
September 25, 2014
3. [20 pts.] Blood plasma is stored at 40◦ F. Before the plasma can be used, it must be at
90◦ F. When the plasma is placed in an oven at 120◦ F, it takes 45 min for the plasma
to warm to 90◦ . How long will it take for the plasma to warm to 90◦ F if the oven
temperature is set at 100◦ F?
dT
= k(A − T ), here A − T > 0. Solve the DE symbolically, i.e. don’t replace
dt
A with a number until after you have found the general solution T (t).
Hints:
Solution:
dT
= k(A − T )
dt
Z
Z
dT
= k dt
u=A−T
A−T
Z
du
−
= kt + C
u
ln |A − T | = −kt + C
− du = dT
A − T = Ce−kt
T (t) = A + Ce−kt
40 = 120 + C =⇒
T (t) = 120 − 80e
C = −80
−kt
Use T (45) = 90 to determine k.
90 = 120 − 80e−45k
3
= e−45k
8
−45k = ln(3/8)
1
k = − ln(3/8) ≈ 0.0217962
45
Finally, solve for t when A = 100, T (0) = 40, and use the k value we just found.
T (t) = 100 + Ce−kt
40 = 100 + C =⇒
90 = 100 − 60e−kt
ln(1/6) = −kt
1
ln(1/6) =
ln(3/8)t
45
ln 6
t = −45
ln(3/8)
t ≈ 82 min
C = −60
Math 2250–016 (Underdown)
Exam 1, Page 4 of 8
September 25, 2014
4. [15 pts.] Use Elimination (you may use Gaussian Elimination on the augmented matrix)
to find the solution set of the following linear system of equations:

 x + 2y + z = 4
3x + 8y + 7z = 20

2x + 7y + 9z = 23
Solution:


1 2 1 4
1 +R2
3 8 7 20 −3R
−→
2 7 9 23


1 2 1 4
1
( )R2
2
0 2 4 8 −→
0 3 7 15


1 2 1 4
0 1 2 4 ⇐⇒
0 0 1 3
(1)
(2)
(3)


1 2 1 4
0 2 4 8
2 7 9 23


1 2 1 4
0 1 2 4
0 3 7 15
−2R1 +R3
−→
−3R2 +R3
−→
x + 2y + z = 4
y + 2z = 4
z=3
y + 2(3) = 4
x + 2(−2) + (3) = 4
The solution set for the system is {(5, −2, 3)}.
⇒
⇒
y = −2
x=5
Math 2250–016 (Underdown)
Exam 1, Page 5 of 8
September 25, 2014
5. [15 pts.] A brine solution of salt flows at a constant rate of 4 L/min into a large tank
that initially holds 100 L of water with 5 kg of salt dissolved in it. The solution inside
the tank is kept well stirred and flows out of the tank at a rate of 4 L/min. If the
concentration of salt in the brine entering the tank is 0.2 kg/L, determine the mass of
salt in the tank after t min. Let x(t) represent the amount of salt in the tank at time t,
measured in kg.
Solution:
dx
= cin fin − cout fout
dt
dx
x
= .2 · 4 −
·4
dt
100
dx
+ .04x = .8
dt
R
I(t) = e .04 dt = e.04t
Z
.04t
xe = .8 e.04t dt
xe.04t = .8(25)e.04t + C
xe.04t = 20e.04t + C
x = 20 + Ce−.04t
Use the initial value x(0) = 5 to determine C.
5 = 20 + Ce0
C = −15
x(t) = 20 − 15e−.04t
Math 2250–016 (Underdown)
Exam 1, Page 6 of 8
September 25, 2014
6. [10 pts.] Use Euler’s method to approximate a solution to the IVP:
dy
=t·y
dt
y(0) = 2,
with a step size of h = 0.5 from t0 = 0 to tf = 2. Recall:
dy
= f (t, y)
y(t0 ) = y0
dt
tn+1 = tn + h
yn+1 = yn + f (tn , yn ) · h
|
{z
}
∆y
Fill in the shaded blanks in the following table:
f (t, y) · h =
n
t
y
∆y
0
0
2
(0 · 2) · 0.5
=
0
1
0.5
2
(0.5 · 2) · 0.5
=
0.5
2
1.0
2.5
(1.0 · 2.5) · 0.5
=
1.25
3
1.5
3.75
(1.5 · 3.75) · 0.5
=
2.8125
4
2.0
6.5625
Math 2250–016 (Underdown)
Exam 1, Page 7 of 8
September 25, 2014
7. [20 pts.] True or False. Circle one.
(a) T
F
A solution to a differential equation must be a differentiable function.
(b) T
F The general solution to a differential equation is actually an infinite family
(set) of solutions.
(c) T
F
If an equation is separable then it is linear.
(d) T
F
If an equation is autonomous then it is separable.
(e) T
F
The equation yy 0 = x is linear.
(f) T
F
The equation y 0 = sin(x)y + 2x is linear.
(g) T
F
The equation y 00 + sin y = 0 is linear.
(h) T
F
The matrix
0 1 7 2 1
0 0 0 0 1
is in reduced row–echelon form (RREF).
(i) T
F
The matrix

2
0

0
0

2
1

0
0
is in row–echelon form (REF).
(j) T
F
If an IVP has a solution then that solution must be unique.
8. [5 points (bonus)] Use geometry to determine a first order differential equation whose
family of solution curves are concentric circles centered on the origin.
Solution: The radial line has slope m = xy . The tangent line to the circle at the
point (x, y) is perpendicular and has slope m⊥ = − xy . Thus,
(x, y)
O
dy
x
=−
dx
y
Z
Z
y dy = −x dx
y2
x2
=− +C
2
2
x2 + y 2 = C
Math 2250–016 (Underdown)
Exam 1, Page 8 of 8
September 25, 2014
Scratch Paper
Question:
1
2
3
4
5
6
7
8
Total
Points:
10
10
20
15
15
10
20
0
100
Bonus Points:
0
0
0
0
0
0
0
5
5
Score: