This preview shows page 1. Sign up to view the full content.
Exercises 4.3
c
2
from the initial conditions.
y
(0) =
−
2=
c
1
+
c
2
,
y
0
(0) = 0 =
−
3
c
1
+
c
2
⇒
c
1
=
−
1
/
2
,
c
2
=
−
3
/
2
.
Thus, the solution to the initial value problem is
y
(
t
)=
−
e
−
3
t
2
−
3
e
t
2
,
and, as
t
→∞
, it approaches
−∞
.
(e)
As in the previous problem, we have negative stiFness
k
=
−
6. But this time the initial
displacement,
y
(0) = 1, as well as the initial velocity,
y
0
(0) = 1, is positive. So, the
mass will start moving in the positive direction, and will continue doing this (due to the
negative stiFness) with increasing velocity. Thus our prediction is that
y
(
t
)
when
t
.
Indeed, the roots of the characteristic equation in this problem are
r
=
−
2 and 3, and
so a general solution has the form
y
(
t
c
1
e
−
2
t
+
c
2
e
3
t
. To satisfy the initial conditions,
we solve the system
y
(0) = 1 =
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

Click to edit the document details