18.03 Muddy
Card
responses,
March
3,
2006
1.
I confused a number of people by dividing through
m
¨
x
+
bx
˙ +
kx
=
F
ext
(
t
) and
¨
miraculously getting
x
+
bx
˙ +
kx
=
q
(
t
).
What I meant to say was that by dividing
through you make the coeﬃcient of
x
equal to 1 and replace the others by
b/m
and
¨
k/m
; but then I’ll change the meaning of the symbol
b
so that it means the damping
constant divided by the mass, and of
k
so that it means the spring constant divided
by the mass.
2.
Here’s the proof that if
x
1
and
x
2
are solutions of the
homogeneous, linear
second
¨
order ODE
x
+
b
(
t
) ˙
x
+
k
(
t
)
x
= 0 then so is
x
=
c
1
x
1
+
c
2
x
2
.
Plug this
x
into the
equation:
k
(
t
)
x
=
k
(
t
)
c
1
x
1
+
k
(
t
)
c
2
x
2
b
(
t
) ˙
x
=
b
(
t
)
c
1
x
˙
1
+
k
(
t
)
c
2
x
˙
2
x
¨
¨
¨
=
c
1
x
1
+
c
2
x
2
¨
x
1
+
b
(
t
) ˙
x
1
+
k
(
t
)
x
1
)
+
c
2
(¨
x
+
b
(
t
) ˙
x
+
k
(
t
)
x
=
c
1
(¨
x
2
+
b
(
t
) ˙
x
2
+
b
(
t
)
x
2
)
and both terms on the right are zero, by assumption.
As the notation indicates, this
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This note was uploaded on 01/31/2011 for the course MAT 17A taught by Professor Staff during the Winter '08 term at UC Davis.
 Winter '08
 Staff

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