Math 2250
Exam 1
June 5, 2014
Name
1. [10 pts.] Find a particular solution to the IVP:
xy 0 = xy + x2 ex
y(0) = 1.
Solution: First put this linear equation in standard form.
y 0 − y = xex
p(x) = −1, thus
R
R
p(x)dx = −x, thus ρ(x) = e p(x)dx = e−x .
Z
Z
−x
x −x
ye = xe e dx = x dx
x2
+C
2 2
x
y=
+ C ex
2
ye−x =
Use the initial value, y(0) = 1 to solve for C:
1 = (0 + C)e0
=⇒
2
x
+ 1 ex
y=
2
C=1
2. [10 pts.] Write the differential equation which corresponds with the tank diagram. Use
x(t) to represent the amount of salt in the tank at time t.
DO NOT SOLVE THE DE!
Solution:
cin
fin
fout
v
v0
= 2 g/l
= 4 l/min
= 3 l/min
= 50 l
= 10 l
dx
= cin · fin − cout · fout
dt
dx
x
= cin · fin −
· fout
dt
v(t)
dx
x
=2·4−
·3
dt
v(t)
v(t) = v0 + (fin − fout )t
= 10 + (4 − 3)t
dx
3
=8−
·x
dt
10 + t
Math 2250
Exam 1, Page 2 of 6
June 5, 2014
3. [20 pts.] Bob has found that he gets peak enjoyment from his cup of coffee when it is
70◦ C. At time t = 0, Bob measured the temperature of his coffee and found it to be 92◦
C. One minute later, Bob measured the coffee again and it was 82◦ C. If the temperature
of the room (the ambient temperature) is 22◦ C, when will Bob’s coffee be 70◦ C?
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
Since A < T , A − T < 0 =⇒ |A − T | = T − A.
T − A = Ce−kt
T (t) = A + Ce−kt
Use the intitial condition, T (0) = 92 to determine C.
92 = 22 + C
T (t) = 22 + 70e
=⇒
C = 70
−kt
Use the second data point, T (1) = 82 to determine k.
82 = 22 + 70e−k
6
= e−k
7
6
−k = ln
7
Finally, solve for t when T = 70.
70 = 22 + 70e−kt
24
= e−kt
35
24
ln
= −kt
35
24
6
ln
= ln
t
35
7
ln 24
35
t=
ln 67
t ≈ 2.45 minutes
− du = dT
Math 2250
Exam 1, Page 3 of 6
June 5, 2014
4. [20 pts.] Use Elimination (you may use Gaussian Elimination) to find the solution set
of the following linear system of equations:

 x + 3y − z = 9
5x + y + z = 23

−x + 2y − 3z = −3
(1)
(2)
(3)
Solution: Let’s use equations (1) and (2) to eliminate x:
−5x + −15y + 5z = −45
5x +
y + z = 23
−14y + 6z = −22
−5(1)
(2)
(10 )
Next, use equations (1) and (3) to eliminate x:
x + 3y − z = 9
−x + 2y − 3z = −3
5y − 4z = 6
(1)
(3)
(20 )
Now we have the 2 × 2 system:
−14y + 6z = −22
5y − 4z = 6
(10 )
(20 )
Let’s combine equations (10 ) and (20 ) to eliminate z:
−28y + 12z = −44
15y − 12z = 18
−13y
= −26
2(10 )
3(20 )
Thus y = 2. Next we’ll backsubstitute this value for y into equation (20 ) to solve for
z:
5(2) − 4z = 6
−4z = −4
z=1
Finally, we’ll backsubstitute these values for y and z into equation (1) to solve for x:
x + 3(2) − 1 = 9
x=4
The solution set for the system is then {(4, 2, 1)}.
Math 2250
Exam 1, Page 4 of 6
June 5, 2014
5. [20 pts.] Initially 5 grams of salt are dissolved in 20 liters of water. Brine with a concentration of 2 grams of salt per liter is added at a rate of 3 liters a minute. The tank
is mixed well and is drained at 3 liters a minute. How long does the process have to
continue until there are 20 grams of salt in the tank?
Solution:
dx
= cin fin − cout fout
dt
dx
x
=2·3−
·3
dt
20
dx
3
+ x=6
dt
20
3
ρ(t) = e 20 t
Z
3
3
t
xe 20 = 6 e 20 t dt
3
3
20
t
e 20 t + C
xe 20 = 6
3
3
3
xe 20 t = 40e 20 t + C
3
x = 40 + Ce− 20 t
Use the initial value x(0) = 5 to determine C.
5 = 40 + Ce0
C = −35
3
x(t) = 40 − 35e− 20 t
Now we solve x(t) = 20 for t.
3
20 = 40 − 35e− 20 t
3
−20 = −35e− 20 t
3
4
= e− 20 t
7
4
3
=− t
ln
7
20
20
4
t = − ln
3
7
t ≈ 3.73 min
Math 2250
Exam 1, Page 5 of 6
June 5, 2014
6. [20 pts.] True or False. Circle one.
(a) T
F
All linear differential equations are separable.
(b) T
F
If a differential equation is nonlinear, then we cannot find an analytic
solution.
x
(c) T
F The differential equation: y 0 =
is linear.
y
(d) T
F
The differential equation: y 0 + sin(t)y = t2
is linear.
(e) T
F If the function y(t) is a solution to a differential equation, then y(t) + C
is also a solution to the same equation for any real value of C.
(f) T
F
If y1 (t) and y2 (t) are solutions to the differential equation:
y 0 + f (t)y = 0,
then 3y1 (t) + 5y2 (t) is also a solution.
(g) T
F
The differential equation: y 00 (t) = t − 5
(h) T
F
The logistic model is an autonomous differential equation.
(i) T
F
The matrix:
is autonomous.


0 1 1 9
0 1 5 4
0 0 0 0
is in row–echelon form.
(j) T F A system of linear algebraic equations with more unknowns than equations
always has an infinite number of solutions.
Math 2250
Exam 1, Page 6 of 6
June 5, 2014
Scratch Paper
Page:
1
2
3
4
5
Total
Points:
20
20
20
20
20
100
Score: