Linear Algebra and Differential Equations - Solution for Problem | MATH 2250, Assignments of Linear Algebra

Material Type: Assignment; Professor: Bornholdt; Class: Linear Algebra and Differential Equations; Subject: Mathematics; University: Utah State University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 07/30/2009

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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
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
.

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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

,sinh( )

cosh( )

x x x x

e e

x

e e

x

 

, we have

x x

x x x x

e

C D

e

C D

A x B x

e e

D

e e

A x B x C

y A x B x C x D x

 

cos( 2 ) sin( 2 )

cos( 2 ) sin( 2 )

cos( 2 ) sin( 2 ) cosh( 2 ) sinh( 2 )

This allows us to determine that the four roots to the corresponding characteristic

equation are

r 2 i

1 , 2



and

2

3 , 4

r 

Roots Factors

r  2 i r  2 i

r  2 i r  2 i

r  2 r  2

r  2 r  2

Therefore, the characteristic equation is

2 2

rr  

or

4

r  .

This means that the original differential equation is

( 4 )

yy