Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil

Solution Manual Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil There is no Solution Manual for Chapter 1 Advanced Mathematics for Engineering Students The Essential Toolbox, 1e Brent J. Lewis, Nihan Onder, Andrew Prudi 1 / 4

Problems Chapter 2 2.1Consider the nonhomogeneous system of rst-order, linear dierential equationsy

1=y2+ coshtandy

2=y1, with boundary conditionsy1(0) = 0 andy2(0) = 1 2 . Solve this initial value problem by the following three

methods:

(a)<> Matrix methods. Use the method of variation of parameters to deter- mine the particular solution.(b)<> Laplace transform methods.(c)<> Convert the two rst-order dierential equations into a single second- order dierential equation and solve this latter equation.Solution (a) y

1=0y1+ 1y2+ cosht y

2=1y1+ 0y2+ 0 Therefore,

y

1 y

2

=

• 1
• 0

y1 y2

+

cosht

and

y1(0) y2(0)

=

1 2

:

Homogeneous equation:y

=

0 11 0

y)

Characteristic equation: det(AI) =

1 1

= 2

1 = 0:

Therefore, 2

= 1)Eigenvalues:1= +1 and2=1.

The eigenvectors are obtained from:x1+x2= 0.

For1= 1)x1=x2and we can take

1 1

:

For2=1)x1=x2and we can take

1 1

:

The homogeneous solution is:y

(h) =c1x (1) e t +c2x (2) e t , where the eigenvec-

tors are:x

(1) =

1 1

andx (2) =

1 1

:

Method of Variation of Parameters fory (p)

:

Y= [y (1) y (2) ] =

e t e t e t e t

andg=

cosht

:

Y 1 = 1 detY

y22y12 y21y11

= 1 2

e t e t e t e t

= 1 2

e t e t e t e t

:

2-1 2 / 4

Check:Y Y

1 = 1 2

e t e t e t e t

e t e t e t e t

=

• 0
• 1

Y 1 g= 1 2

e t e t e t e t

e t +e t 2

= 1 4

• +e

2t

• +e

2t

:

Therefore,u(t) = 1 4 Z t

• +e

2^t

• +e

2^t

d^t= 1 4 2 6 6 4 t+

e 2t 1 2

t+

e 2t 1 2

3 7 7 5

:

y (p) =Y u= 1 4

e t e t e t e t

2 6 6 4 t+

e 2t 1 2

t+

e 2t 1 2

3 7 7 5 = 1 4 2 6 4 te t

e t 2 + e t 2 +te t + e t 2

e t 2 te t

e t 2 + e t 2 te t

e t 2 + e t 2 3 7 5= 1 2

tcosht+ sinht tsinht

:

The last expression is obtained after factoring. The general solution is thus:

y (g) =c1

1 1

e t +c2

1 1

e t + 1 2

tcosht+ sinht tsinht

Now,

y1(0) y2(0)

=

1 2

so that:

y1(0) y2(0)

=c1

11

+c2

1 1

+

00

=

1 2

which yields:

c1+c2= 0 c1c2= 1 2

:

Thus,c1= 1 4 andc2= 1 4 :Therefore: y1(t) = 1 4 e t + 1 4 e t |{z} =

sinht 2 + t 2 cosht+

sinht 2 = t 2 cosht y2(t) = 1 4 e t

1 4 e t |{z} = cosht 2 + t 2 sinht= 1 2 [tsinhtcosht][answer]

2-2 3 / 4

(b) Using a Laplace Transform method:

sY1=Y2+ s s 2 1 )Y2=sY1 s s 2 1 sY2+ 1 2 =Y1)sY2=Y1 1 2 (2.1)

Multiplying the rst equation through by (s):

sY2=s 2 Y1+ s 2 s 2 1 sY2=Y1 1 2

Adding the above two equations and solving forY1:

Y1= s 2 (s 2 1) 2

1 2 1 (s 2 1)

Taking the inverse Laplace transform ofY1:

y1(t) =L 1 fY1g= sinht+tcosht 2

1 2 sinht= tcosht 2 [answer].

Rewriting eq. (2.1):

sY1=Y2+ s s 2 1 Y1=sY2+ 1 2

Multiplying the second equation through by (s):

sY1=Y2+ s s 2 1 sY1=s 2 Y2 s 2

Adding the above two equations and solving forY2:

Y2= s (s 2 1) 2

s 2(s 2 1)

Taking the inverse Laplace transform ofY2:

y2(t) =L 1 fY2g= tsinht 2

1 2 cosht[answer].The solutions fory1(t) andy2(t) are identical to those in part (a).2-3

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