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 =
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Constant of integration, Boundary value problem

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