California State University Northridge
MATH 280: Applied Differential Equations
Midterm Exam 2
November 13, 2013. Duration: 75 Minutes. Instructor: Jing Li
Student Name:
Student number:
Take your time to read the entire paper before you begin to write, and read each question
carefully. Remember that certain questions are worth more points than others. Make a note of
the questions that you feel confident you can do, and then do those first: you do not have to
proceed through the paper in the order given.
• You have 75 minutes to complete this exam.
• This is a closed book exam, and no notes of any kind are allowed. The use of cell phones,
pagers or any text storage or communication device is not permitted.
• Only the Faculty approved TI-30 calculator is allowed.
• The correct answer requires justification written legibly and logically: you must convince
me that you know why your solution is correct. Answer these questions in the space provided. Use the backs of pages if necessary.
• Where it is possible to check your work, do so.
• Good Luck!
Problem
1
2
3
4
5
6
7
Points
8
24
24
12
30
30
20
Your Marks
1
8
(BONUS)
24
Total
148
Question 1. [8 points] Determine whether the given set of fuctions is linearly independent on
the interval (−∞, ∞).
(a) [4 points] f 1 (x) = x, f 2 (x) = x 2 , f 3 (x) = 4x − 3x 2
Solution: Since
f 3 (x) = 4 · f 1 (x) − 3 · f 2 (x) =⇒ 4 · f 1 (x) − 3 · f 2 (x) − f 3 (x) = 0,
the set of functions f 1 (x), f 2 (x), f 3 (x) is linearly dependent.
(b) [4 points] f 1 (x) = 1 + x, f 2 (x) = x, f 3 (x) = x 2
Solution: Let
c 1 f 1 (x) + c 2 f 2 (x) + c 3 f 3 (x) = 0 =⇒ c 1 (1 + x) + c 2 x + c 3 x 2 = 0 =⇒ c 1 + (c 1 + c 2 )x + c 3 x 2 = 0,
then c 1 = 0, c 1 + c 2 = 0, c 3 = 0, =⇒ c 1 = c 2 = c 3 = 0.
Hence the set of functions f 1 (x), f 2 (x), f 3 (x) is linearly independent.
2
Question 2. [24 points] Consider the differential equation
x 3 y ��� + x 2 y �� − 2x y � + 2y = 0
(a) [15 points] Verify that the fuctions of y 1 = x, y 2 = x 2 , y 3 =
differential equation.
1
x
are solutions of the given
Solution:
• Check y 1 is the sol of x 3 y ��� + x 2 y �� − 2x y � + 2y = 0
y 1 = x =⇒ y 1� = 1, y 1�� = 0, y 1��� = 0
Then
x 3 y 1��� + x 2 y 1�� − 2x y 1� + 2y 1 = 0 + 0 − 2x + 2x = 0.
• Check y 2 is the solution of x 3 y ��� + x 2 y �� − 2x y � + 2y = 0
y 2 = x 2 =⇒ y 2� = 2x, y 2�� = 2, y 2��� = 0
Then
x 3 y 2��� + x 2 y 2�� − 2x y 2� + 2y 2 = 0 + x 2 · 2 − 2x · 2x + 2x 2 = 0.
• Check y 3 is the solution of x 3 y ��� + x 2 y �� − 2x y � + 2y = 0
y3 =
1
=⇒ y 3� = −x −2 , y 3�� = 2x −3 , y 3��� = −6x −4
x
Then
x 3 y 3��� + x 2 y 3�� − 2x y 3� + 2y 3 = 0.
(b) [7 points] Verify that y 1 , y 2 , y 3 in part (a) form a fundamental set of solutions for the
given differential equaiton on the interval of (0, ∞).
Solution:
�
� y1
�
W (y 1 , y 2 , y 3 )(x) = �� y 1�
� y ��
1
y2
y 2�
y 2��
y3
y 3�
y 3��
� �
1
� � x x2
x
� �
� = � 1 2x −x −2
� �
� � 0 2 2x −3
Then y 1 , y 2 , y 3 form a fundamental set of solutions.
�
�
�
� = 6 1 �= 0, on (0, ∞)
�
x
�
(c) [2 points] Using y 1 , y 2 and y 3 in part (a), form the general solution of the given differential
equation.
Solution: The general solution is y(x) = c 1 y 1 + c 2 y 2 + c 3 y 3 = c 1 x + c 2 x 2 + c 3 x1 .
3
Question 3. [24 points] Find the general solution of the given second-order differential equation.
(a) [8 points] y �� − 7y � + 12y = 0
Solution: m 2 − 7m + 12 = 0 =⇒ (m − 3)(m − 4) = 0 =⇒ m 1 = 3, m 2 = 4 =⇒ y = c 1 e 3x + c 2 e 4x .
(b) [8 points] y �� − 6y � + 9y = 0
Solution: m 2 − 6m + 9 = 0 =⇒ (m − 3)2 = 0 =⇒ m 1 = m 2 = 3 =⇒ y = c 1 e 3x + c 2 xe 3x .
(c) [8 points] y �� − 4y � + 5y = 0
Solution: m 2 − 4m + 5 = 0 =⇒ m 1,2 = 2 ± i =⇒ y = e 2x (c 1 cos x + c 2 sin x).
4
Question 4. [12 points] Find the general solution of the given third-order differential equation
d 3x d 2x
+
− 2x = 0
dt3 dt2
Solution: m 3 + m 2 − 2 = 0 =⇒ m = 1is one of the roots, then using Long Division, we have
m 3 + m 2 − 2 = (m − 1)(m 2 + 2m + 2)
Then m 2,3 = −1 ± i .
Hence, the general solution is
x = c 1 e t + e −t (c 1 cos t + c 2 sin t )
5
Question 5. [30 points] Solve the given differential equation by undetermined coefficients,
y �� + 2y � = 2x + 5 − e −2x
subject to the initial conditions y(0) = 0, y � (0) = 12 .
Solution:
The associated homogeneous equation is y �� + 2y � = 0.
Then the auxiliary equation is m 2 + 2m = 0 =⇒ m = 0andm = 2.
Hence, the complementary function is y c = c 1 e −2x + c 2 .
Now g (x) = 2x + 5 − e −2x =⇒ y p (x) = Ax + B +C e −2x , however, there is duplications between
y c and y p . Then the correct assumed particular solution is y p (x) = Ax 2 + B x +C xe −2x .
Then y p� (x) = 2Ax + B +C e −2x − 2C xe −2x and y p�� (x) = 2A − 4C e −2x + 4C xe −2x .
Hence, y p�� (x) + 2y p� (x) = 2A + B + 4Ax − 2C e −2x = 2x + 5 − e −2x =⇒ 2A + 2B = 5, 4A = 2, −2C =
−1 =⇒ A = 12 , B = 2,C = 12 =⇒ y p (x) = 12 x 2 + 2x + 21 xe −2x
Then the general solution is
1
1
y = c 1 e −2x + c 2 + x 2 + 2x + xe −2x
2
2
Using I.C. =⇒ 0 = c 1 + c 2 , 12 = −2c 1 + 2 + 12 =⇒ c 1 = 1, c 2 = −1.
Therefore, the solution is
1
1
y = e −2x − 1 + x 2 + 2x + xe −2x
2
2
6
Question 6. [30 points] Solve the given differential equation by variation of parameters
3y �� − 6y � + 6y = e x sec x
Solution:
1. For 3y �� − 6y � + 6y = 0 we have 3m 2 − 6m + 6 = 0 ⇒ 3(m 2 − 2m + 2) = 0. So
�
2 ± 4 − 4 · 2 2 ± 2i
m1 , m2 =
=
= 1±i
2
2
and α = 1 = β. So, y c = e x (c 1 cosx + c 2 sin x). Hence y 1 = e x cos x and y 2 = e x sin x.
2. We have
�
� y1
w(y 1 , y 2 ) = �� �
y1
� �
y 2 �� ��
e x cos x
e x sin x
� �=� x
x
x
y2
e cos x − e sin x e sin x + e x cos x
�
�
�
�
= e x cos x(e x sin x + e x cos x) − e x sin x(e x cos x − e x sin x)
= e 2x cos x sin x + e 2x cos2 x − e 2x sin x cos x + e 2x sin2 x
= e 2x (cos2 x + sin2 x) = e 2x .
3. 3y �� − 6y � + 6y = e x sec x ⇒ y �� − 2y � + 2y = 13 e x sec x. f (x) = 13 e x sec x.
4. We have
So
�
� 0
w 1 = ��
f (x)
�
� y1
w 2 = �� �
y1
� �
�
�
0
e x sin x
y 2 �� ��
� = − 1 e 2x sin x sec x
� �=� 1 x
x
x
�
y2
e
sec
x
e
sin
x
+
e
cos
x
3
3
� �
�
x
�
�
�
e cos x
0
0 � �
� = 1 e 2x cos x sec x.
=� x
1 x
x
�
f (x)
e cos x − e sin x
e sec x � 3
3
1 2x
w 1 − 3 e sin x sec x
1
1
1
1
=
=
= − sin x sec x = − sin x
= − tan x.
2x
w
e
3
3
cos x
3
1 2x
e
cos
x
sec
x
w2 3
1
1
u 2� =
=
= cos x sec x = .
2x
w
e
3
3
u 1�
Therefore,
�
�
�
�
1
1 sin x
1 sin x
1
1
1
u 1 = − tan xd x = −
dx = −
dx =
d (cos x) = ln | cos x|.
3 cos x
3 cos x
3 cos x
3
� 3
1
1
u2 =
d x = x.
3
3
5. y p = u 1 y 1 + u 2 y 2 = 13 e x cos x ln | cos x| + 13 xe x sin x.
6. y = e x (c 1 cosx + c 2 sin x) + 13 e x cos x ln | cos x| + 13 xe x sin x.
7
Question 7. [20 pionts] A 100-volt electromotive force is applied to an RC-series circuit in
which the resistance is 200 ohms and the capacitance is 10−4 farad.
(a) [16 points] Find the charge q(t ) on the capacitor if q(0) = 0.
dq
Solution: Assume that R d t + C1 q = E (t ) where R = 200 ohms, C = 10−4 farad, and E (t ) = 100.
Then
dq
dq
dq
1
200 d t + 104 q = 100 =⇒ 2 d t + 100q = 1 =⇒ d t + 50q = 12 =⇒ q = 100
+ ce −50t
1
Using I.C., we get c = − 100
. Hence the solution is
q=
1
1 −50t
−
e
.
100 100
(b) [4 points] Find the current i (t ).
Solution: i =
dq
dt
1
= − 100
· (−50)e −50t = 12 e −50t .
8
Question 8. [BONUS: 24 points] Find the general solution of
x 4 y �� + x 3 y � − 4x 2 y = 1
given that y 1 = x 2 is a solution of the associated homogeneous equation.
[HINT: Reduction of Order & Variation of Parameters]
Solution: We are given that y 1 = x 2 is a sol. of x 4 y �� + x 3 y � − 4x 2 y = 1.
To find a second solution, we use reduction of order. Let y = x 2 u(x). Then the product rule
gives y � = x 2 u � + 2xu and y �� = x 2 u �� + 4xu � + 2n. So
x 4 y �� + x 3 y � − 4x 2 y = x 5 (xu �� + 5u � ) = 0
Letting w = u � , this becomes xw � + 5w = 0
Separating variables and integrating we have
dw
5
= − d x and ln |w| = −5 ln x +C
dt
x
Thus, w = x −5 and u = − 14 x −4 .
A second solution is then y 2 = x 2 · x −4 = x12 , and the general solution of the homogeneous
DE is y c = c 1 x 2 + xc22 .
To find a particular solution, y p , we use variation of parameters. The Wronskian is W = − x4 .
Identifying f (x) = x14 , we obtain
1 −4
x
u 1� = 14 x −5 =⇒ u 1 = − 16
1 −1
1
�
u 2 = − 4 x =⇒ u 2 = − 4 ln x
1 −4 2
1 −2
So y p = − 16
x x − 14 (ln x)x −2 = − 16
x − 14 x −2 ln x.
The general solution is
c2
1
1
y = c 1 x 2 + 2 − x 2 − 2 ln x.
x
16
4x
9