MATH 2250 Linear Algebra/Differential Equations
5.3 Problem 41 The crux of this problem is unraveling the hyperbolic functions. We are
given the solution to a linear homogeneous differential equation and asked to determine
the differential equation itself.
Using
2
)sinh(,
2
)cosh(
xxxx
ee
x
ee
x
, we have
xx
xxxx
e
DC
e
DC
xBxA
ee
D
ee
CxBxA
xDxCxBxAy
22
)2sin()2cos(
22
)2sin()2cos(
)2sinh()2cosh()2sin()2cos(
This allows us to determine that the four roots to the corresponding characteristic
equation are
ir 2
2,1
and
2
4,3
r
.
Roots Factors
ir 2
ir 2
ir 2
ir 2
2r
2r
2r
2r
Therefore, the characteristic equation is
0)4)(4(
22
rr
or
016
4
r
.
This means that the original differential equation is
016
)4(
yy
.