University of California, Berkeley
ME 180, Engineering Analysis Using the Finite Element Method
Spring 2008
Instructor: T. Zohdi
Quiz 1 Solutions
Problem 1
Solve
d
dx
A
1
du
dx
!
=
k
2
sin
2
π
k
L
x
!
,
analytically, where
A
1
,
k
, and
L
are constants.
Solution:
(5 points) We know that the analytical solution for any linear, nonhomogeneous ODE
can be written as
u
(
x
)
=
u
h
(
x
)
+
u
p
(
x
)
,
i.e., it can be broken up into homogeneous and particular parts. The equation corresponding to
the homogeneous part of the solution is
A
1
d
2
u
dx
2
=
0
.
Considering the proposed solution
u
=
e
λ
x
results in the algebraic equation
A
1
λ
2
e
λ
x
=
0
.
Assuming that
A
1
,
0, and knowing that the exponential function never equals zero, results in
λ
1
=
λ
2
=
0, i.e., repeated real roots. This results in a homogeneous solution
u
h
=
c
1
xe
0
x
+
c
2
e
0
x
=
c
1
x
+
c
2
.
The particular solution can be solved various ways (undetermined coe
ﬃ
cients, variation of pa
rameters). If one considers a proposed particular solution
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 Spring '08
 ZOHDI
 Finite Element Method, dx, total solution, dx dx

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