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ECE220
Solution Lab #6
Review for Test 2
Date: Week of March 10, 2008
The goal of this lab is to review material in preparation for Test 2.
Work out as many problems as you can. If you have worked on a problem before,
you may skip it. Do not worry if you do not ﬁnish all problems in time.
I. Classiﬁcation of diﬀerential equations
1. Give two examples of the following:
•
A linear DE
(a)
d
2
v
dt
2
+ 2
dv
dt
+ 6
v
(
t
) = 5
e

2
t
u
(
t
)
(b)
d
4
v
dt
4
+
dv
dt
+ 5
tv
(
t
) = 10cos(2
π
250
t
)
u
(
t
)
•
A nonlinear DE
(a) (
d
2
v
dt
2
)
2
+ 2
dv
dt
= 10
u
(
t
)
(b) (
d
3
v
dt
3
)
3
+ 5
v
(
t
) = 10
u
(
t
)
•
A DE of 3rd order
(a)
d
3
v
dt
3
+ 2
d
2
v
dt
2
+
tv
(
t
) =
e

t
u
(
t
)
(b)
d
3
v
dt
3
+
v
(
t
) = 10 cos(
t
)
u
(
t
)
•
A DE with constant coeﬃcients
(a)
d
2
v
dt
2
+ 2
dv
dt
+ 6
v
(
t
) = 5
e

2
t
u
(
t
)
(b)
d
4
v
dt
4
+
dv
dt
+ 5
v
(
t
) = 10 cos(2
π
250
t
)
u
(
t
)
•
A DE with two timevarying coeﬃcients
(a)
t
(
dv
dt
)
2
+
t
4
v
(
t
) = 20
u
(
t
)
(b)
d
3
v
dt
3
+ 2
t
dv
dt
+ 7
tv
(
t
) =
e

12
t
u
(
t
)
1
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View Full Document•
A homogeneous DE
(a)
d
2
v
dt
2
+ 3
dv
dt
+ 12
v
(
t
) = 0
(b)
d
4
v
dt
4
+
v
(
t
) = 0
•
A system of DEs
(a) 2
dv
1
dt
+ 6
v
1
(
t
) + 12
v
2
(
t
) = 5
u
(
t
)
dv
2
dt
+ 5
v
1
(
t
) = sin(2
π
250
t
)
u
(
t
)
(b) 2
dv
1
dt
+ 6
v
1
(
t
) + 12
v
2
(
t
) +
v
3
(
t
) = 5
u
(
t
)
dv
2
dt
+ 5
v
1
(
t
) = sin(2
π
250
t
)
u
(
t
)
dv
3
dt
+
v
2
(
t
) = 50
u
(
t
)
•
A linear DE with constant coeﬃcients
(a)
d
2
v
dt
2
+ 2
dv
dt
+ 6
v
(
t
) = 5
e

2
t
u
(
t
)
(b)
d
4
v
dt
4
+ 4
dv
dt
+ 5
v
(
t
) = 10 cos(2
π
250
t
)
u
(
t
)
————————————————————————————–
II. Linearity and timeshifting properties
Question 1.
Using the results of Examples 7.13, 7.14 and 7.15 (or theorem 7.4, page 348),
determine the particular solution of the DE
dv
(
t
)
dt
+ 4
v
(
t
) = 10
u
(
t
) + 5cos(2
πt
)
u
(
t
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 Spring '08
 NILSON

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