California State University Northridge
MATH 280: Applied Differential Equations
Midterm Exam 2
April 3, 2013. Duration: 75 Minutes. Instructor: Jing Li
Solutions.
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1
Student number:
Problem
1
2
3
4
5
6
7
8
Total
Points
6
16
21
10
25
20
22
21
BONUS
120
Your Marks
Question 1. [6 points] Determine whether the given set of fuctions is linearly independent on
the interval (−∞, ∞).
(a) [3 points] f 1 (x) = x, f 2 (x) = x − 1, f 3 (x) = x + 3
Solution:
Since (−4)x + 3(x − 1) + 1(x + 3) = 0 the set of functions is linearly dependent.
(b) [3 points] f 1 (x) = 2 + x, f 2 (x) = 2 + |x|
Solution:
From the graphs of f 1 (x) = 2 + x and f 2 (x) = 2 + |x| we see that the set of functions is linearly
independent since they cannot be multiples of each other.
The graph of f 2 (x) = 2 + |x| is
The graph of f 1 (x) = 2 + x is
2
Question 2. [16 points] For the differential equation
4y 00 − 4y 0 + y = 0,
(a) [14 points] verify the fuctions of y 1 = e x/2 , y 2 = xe x/2 form a fundamental set of solutions
for the above differential equaiton on the interval of (−∞, ∞);
Solution:
• Check that y 1 is a solution of 4y 00 − 4y 0 + y = 0.
x
x
x
y 1 = e 2 , y 10 = 21 e 2 and y 100 = 14 e 2 .
x
x
x
x
1 x
1 x
4y 100 − 4y 10 + y 1 = 4 · e 2 − 4 · e 2 + e 2 = e 2 − 2e 2 + e 2 = 0.
4
2
• Check that y 2 is a solution of 4y 00 − 4y 0 + y = 0.
x
x
x
x
x
x
x
x
y 2 = xe 2 , y 20 = e 2 + 21 xe 2 and y 200 = 12 e 2 + 21 e 2 + 14 xe 2 = e 2 + 41 xe 2 .
x
x
x
x
x
x
x
x
1 x
1 x
4y 200 − 4y 20 + y 2 = 4(e 2 + xe 2 ) − 4(e 2 + xe 2 ) + xe 2 = 4e 2 + xe 2 − 4e 2 − 2xe 2 + xe 2 = 0.
4
2
• Check y 1 and y 2 are linearly independent.
¯
¯ y1
w(y 1 , y 2 ) = ¯¯ 0
y1
¯ ¯ x
y 2 ¯¯ ¯¯ e 2
=
x
y 20 ¯ ¯ 12 e 2
¯
x
¯
x
x
x
x x
1 x x
x
xe 2
e 2 ) − xe 2 e 2 = e x + e x − e x = e x 6= 0
x
x ¯ = e 2 (e 2 +
x
¯
e 2 + 2e 2
2
2
2
2
on (−∞, ∞). Therefore, y 1 and y 2 are linearly independent.
Hence y 1 and y 2 form a fundamental set of solutions for the above DE on (−∞, ∞).
(b) [2 points]using y 1 and y 2 in part (a), form the general solution of the above differential
equation.
x
x
Solution: y = c 1 y 1 + c 2 y 2 = c 1 e 2 + c 2 xe 2 .
3
Question 3. [21 points] Find the general solution of the given second-order differential equation.
(a) [7 points] y 00 − 3y 0 + 2y = 0
Solution: From m 2 −3m+2 = 0 we have (m−2)(m−1). So m 1 = 1 and m 2 = 2. So y = c 1 e x +c 2 e 2x .
(b) [7 points] y 00 − 10y 0 + 25y = 0
Solution: From m 2 − 10m + 25 = 0 we have (m − 5)2 = 0. So m 1 = m 2 = 5 and y = c 1 e 5x + c 2 xe 5x .
(c) [7 points] 2y 00 + 2y 0 + y = 0
Solution: From 2m 2 + 2m + 1 = 0 we have that
p
p
1 1
−2 ± 22 − 4 · 2 · 1 −2 ± 4 − 8 −2 ± 2i
=
=
= − ± i.
m1 , m2 =
2·2
4
4
2 2
¢
x¡
So α = − 12 and β = 12 . Therefore, y = e − 2 c 1 cos 12 x + c 2 sin 12 x .
4
Question 4. [10 points] Find the general solution of the given third-order differential equation
d 3x d 2x
−
− 4x = 0
dt3 dt2
Solution:
From m 3 − m 2 − 4 = 0 we have that since a 0 = −4 it has factors of ±4, ±2, ±1 and since a 3 = 1
it has factors of ±1.
p
Check m = q , where p is the factor of a 0 and q is the factor of a 3 . We find that m = 2 is one
root.
Now, using long division, we know that m 3 −m 2 −4 = 0 ⇔ (m −2)(m 2 +m +2) = 0. Therefore,
m 1 = 2 or m 2 + m + 2 = 0 in which case
p
p
−1 ± 1 − 4 · 2 −1 ± 7i
=
m2 , m3 =
2
2
and α = − 21 and β =
p
7
2 .
2t
Therefore, x = c 1 e
t
+ e − 2 (c 2 cos
p
7
2 t
+ c 3 sin
p
7
2 t ).
5
Question 5. [25 points] Solve the given differential equation by undetermined coefficients
y 00 − 4y = (x 2 − 3) sin 2x
Solution:
For y 00 −4y = 0, we have m 2 −4 ⇒ m 2 = 4 ⇒ m 1 = 2, m 2 = −2. Then y c = c 1 e 2x +c 2 e −2x . Since
g (x) = (x 2 − e) sin 2x we assume that y p = (Ax 2 + B x +C ) cos 2x + (E x 2 + F x +G) sin 2x. Now,
y p0 = (2Ax + B ) cos 2x + (Ax 2 + B x +C )(−2 sin 2x) + (2E x + F ) sin 2x + (E x 2 + F x +G)(2 cos 2x)
= (2Ax + B ) cos 2x + (−2)(Ax 2 + B x +C ) sin 2x + (2E x + F ) sin 2x + 2(E x 2 + F x +G) cos 2x.
y p00 = 2A cos 2x + (2Ax + B )(−2 sin 2x)
+ (−2)(2Ax + B ) sin 2x + (−2)(Ax 2 + B x +C )(2 cos 2x)
+ 2E sin 2x + (2E x + F )(2 cos 2x) + 2(2E x + F ) cos 2x + 2(E x 2 + F x +G)(−2 sin 2x).
From y p00 − 4y p = (x 2 − 3) sin 2x we have:
For x 2 cos 2x: −4A − 4A = −8A = 0 ⇒ A = 0.
For x cos 2x: −4B + 4E + 4E − 4B = −8B + 8E = 0 ⇒ B = E = − 18 .
For cos 2x: 2A − 4C + 2F + 2F − 4C = 2A − 8C + 4F = 0 ⇒ C = 0.
For x 2 sin 2x: −4E − 4E = −8E = 1 ⇒ E = − 18 .
For x sin 2x: −4A − 4A − 4F − 4F = −8A − 8F = 0 ⇒ F = 0.
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For sin 2x: −2B − 2B + 2E − 4G − 4G = −4B + 2E − 8G = −3 ⇒ G = 32
.
) sin 2x. Therefore, the general solution is y = c 1 e 2x +c 2 e −2x −
Hence y p = − 18 x cos 2x +(− 81 x 2 + 13
32
1
1 2
13
x cos 2x + (− 8 x + 32 ) sin 2x.
8
6
Question 6. [20 points] Solve the given initial-value problem
5y 00 + y 0 = −6x, y(0) = 0, y 0 (0) = −10
Solution:
For 5y 00 + y 0 = 0 we have 5m 2 + m = 0 ⇒ m(5m + 1) = 0 ⇒ m 1 = 0, m 2 = − 15 .
x
So y c = c 1 + c 2 e − 5 .
Since g (x) = −6x we first assume y p = Ax + B . However, B here is duplicated in y c . Then the
correct assumption for y p is y p = Ax 2 + B x. So, y p0 − 2Ax + B and y p00 = 2A.
So 5y p00 + y p0 = 5 · 2A + 2Ax + B = 2Ax + (10A + B ). We have, 2Ax + (10A + B ) = −6x. Therefore
2A = −6 ⇒ A = −3 and 10A + B = 0 ⇒ B = 30. Hence, y p = −3x 2 + 30x.
x
The general solution is y = c 1 + c 2 e − 5 − 3x 2 + 30x. Now applying I.C. to y and y 0 gives y 0 =
x
− 51 c 2 e − 5 − 6x + 30.
y(0) = 0 ⇒ 0 = c 1 + c 2 e 0 ⇒ c 1 + c 2 = 0 ⇒ c 1 = −200.
y 0 (0) = −10 ⇒ −10 = − 51 c 2 e 0 − 0 + 30 ⇒ c 2 = 200.
x
Therefore, the solution of the initial value problem is y = −200 + 200e − 5 − 3x 2 + 30x.
7
Question 7. [22 points] Solve the given differential equation by variation of parameters
3y 00 − 6y 0 + 6y = e x sec x
Solution:
1. For 3y 00 − 6y 0 + 6y = 0 we have 3m 2 − 6m + 6 = 0 ⇒ 3(m 2 − 2m + 2) = 0. So
p
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 ) = ¯¯ 0
y1
¯ ¯
y 2 ¯¯ ¯¯
e x cos x
e x sin x
=
y 20 ¯ ¯ e x cos x − e x sin x e x 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 00 − 6y 0 + 6y = e x sec x ⇒ y 00 − 2y 0 + 2y = 13 e x sec x. f (x) = 31 e x sec x.
4. We have
¯
¯ 0
w 1 = ¯¯
f (x)
¯
¯ y1
w 2 = ¯¯ 0
y1
¯ ¯
¯
¯
0
e x sin x
y 2 ¯¯ ¯¯
¯ = − 1 e 2x sin x sec x
0 ¯=¯ 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
So
1 2x
w 1 − 3 e sin x sec x
1
1
1
1
=
= − tan x.
= − sin x sec x = − sin x
2x
w
e
3
3
cos x
3
1 2x
w 2 3 e cos x sec x 1
1
u 20 =
=
= cos x sec x = .
2x
w
e
3
3
u 10 =
Therefore,
Z
Z
Z
Z
1
1
1 sin x
1 sin x
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
Z 3
1
1
d x = x.
u2 =
3
3
5. y p = u 1 y 1 + u 2 y 2 = 31 e x cos x ln | cos x| + 31 xe x sin x.
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Question 8. [BONUS: 21 points] Discuss how the methods of undetermined coefficients and
variation of parameters can be combined to solve the given differential equation. Carry out
your idea.
y 00 − 2y 0 + y = 4x 2 − 3 + x −1 e x
[HINT: Superposition Principle – Nonhomogeneous Equations]
Solution:
For y 00 − 2y 0 + y = 0 the auxiliary equation is m 2 − 2m + 1 − 0 ⇒ (m − 1)2 = 0 ⇒ m 1 = m 2 = 1.
So, y c = c 1 e x + c 2 xe x .
Now, we consider first the DE y 00 −2y 0 + y = 4x 2 −3 which can be solved using undetermined
coefficients where g 1 (x) = 4x 2 − 3. Then y p 1 = Ax 2 + B x + C and then substituting into the DE
we get
Ax 2 + (−4A + B )x + (2A − 2B +C ) = 4x 2 − 3.
Then A = 4, −4A + B = 0 and 2A − 2B + C = −3. SO A = 4, B = 16 AND C = 21. Thus y p 1 =
4x 2 + 16x + 21.
Next, we consider the DE y 00 − 2y 0 + y = x −1 e x for which a particular solution y p 2 can be
x
found using variation of parameters. The Wronskian is W = e 2x . Identifying f (x) = ex we obtain
u 10 = −1 and u 20 = x1 . Then u 1 = −x and u 2 = ln x so that y p 2 = −xe x + xe x ln x. The general
solution of the original DE is then
y = yc + y p1 + y p2
= c 1 e x + c 2 xe x + 4x 2 + 16x + 21 − xe x + xe x ln x
= c 1 e x + c 3 xe x + 4x 2 + 16x + 21 + xe x ln x.
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