Higher Engineering Mathematics, Sixth Edition

522 Higher Engineering Mathematics

from which,

X′′−μX=0andT′′−μT= 0.

4. Lettingμ=−p^2 to give an oscillatory solution
   gives:

X′′+p^2 X=0andT′′+p^2 T= 0

The auxiliary equation foreach is:m^2 +p^2 = 0

from which,m=

√

−p^2 =±jp.

5. Solving each equation gives:
   X=Acospx+Bsinpx,and
   T=Ccospt+Dsinpt.
   Thus,
   u(x,t)={Acospx+Bsinpx}{Ccospt+Dsinpt}.


6. Applying the boundary conditions to determine
   constantsAandBgives:
   (i) u( 0 ,t)=0,hence 0=A{Ccospt+Dsinpt}
   from which we conclude thatA=0.
   Therefore,
   u(x,t)=Bsinpx{Ccospt+Dsinpt} (a)

(ii) u( 50 ,t)=0, hence

0 =Bsin50p{Ccospt+Dsinpt}. B=0,

hence sin50p=0 from which, 50p=nπand

p=

nπ

50

7. Substituting in equation (a) gives:

u(x,t)=Bsin

nπx

50

{

Ccos

nπt

50

+Dsin

nπt

50

}

or, more generally,

un(x,t)=

∑∞

n= 1

sin

nπx

50

{

Ancos

nπt

50

+Bnsin

nπt

50

}

(b)

whereAn=BCandBn=BD.

8. From equation (8),

An=

2

L

∫L

0

f(x)sin

nπx

L

dx

=

2

50

[∫

25

0

(

2

25

x

)

sin

nπx

50

dx

+

∫ 50

25

(

100 − 2 x

25

)

sin

nπx

50

dx

]

Each integral is determined using integration by

parts (see Chapter 43, page 420) with the result:

An=

16

n^2 π^2

sin

nπ

2

From equation (9),

Bn=

2

cnπ

∫L

0

g(x)sin

nπx

L

dx

[

∂u

∂t

]

t= 0

= 0 =g(x)thus,Bn= 0

Substituting into equation (b) gives:

un(x,t)=

∑∞

n= 1

sin

nπx

50

{

Ancos

nπt

50

+Bnsin

nπt

50

}

=

∑∞

n= 1

sin

nπx

50

{

16

n^2 π^2

sin

nπ

2

cos

nπt

50

+( 0 )sin

nπt

50

}

Hence,

u(x,t)=

16

π^2

∑∞

n= 1

1

n^2

sin

nπx

50

sin

nπ

2

cos

nπt

50

For stretched stringproblems as in problem 5 above, the

main parts of the procedure are:

1. DetermineAnfrom equation (8).

Note that

2

L

∫ L

0

f(x)sin

nπx

L

dxisalwaysequal

to

8 d

n^2 π^2

sin

nπ

2

(see Fig. 53.3)

2. DetermineBnfrom equation (9)


3. Substitute in equation (5) to determineu(x,t)

0 x

y 5 f (x )

L

d

y

L

2

Figure 53.3