Higher Engineering Mathematics, Sixth Edition

An introduction to partial differential equations 519

Worked Problem 4 will be a reminder of solving
ordinary differential equations of this type.

Problem 4. Find the general solution of the

following differential equations:

(a)X′′− 4 X=0(b)T′′+ 4 T=0.

(a) IfX′′− 4 X=0 then the auxiliary equation (see

Chapter 50) is:

m^2 − 4 =0i.e.m^2 =4 from which,

m=+2orm=− 2

Thus, the general solution is:

X=Ae^2 x+Be−^2 x

(b) IfT′′+ 4 T=0 then the auxiliary equation is:

m^2 + 4 =0i.e.m^2 =−4 from which,

m=

√

− 4 =±j 2

Thus, the general solution is:

T=e^0 {Acos 2t+Bsin2t}=Acos2t+Bsin2t

Now try the following exercise

Exercise 200 Further problemson revising

the solution of ordinary differential equation

1. SolveT′′=c^2 μTgivenc=3andμ=1.
   [T=Ae^3 t+Be−^3 t]


2. SolveT′′−c^2 μT=0givenc=3andμ=−1.
   [T=Acos 3t+Bsin3t]


3. SolveX′′=μXgivenμ=1.[
   X=Aex+Be−x

]

4. SolveX′′−μX=0givenμ=−1.
   [X=Acosx+Bsinx]

53.6 The wave equation

Anelastic stringis a string with elastic properties, i.e.
the string satisfies Hooke’s law. Figure 53.1 shows a
flexible elastic string stretched between two points at
x=0andx=Lwith uniform tensionT. The string will
vibrate if the string is displace slightly from its initial
position of rest and released, the end points remaining

0

x

x

y

P

L

u(x, t )

u

5

f^ (

x, t

)

Figure 53.1

fixed. The positionof any pointPon the string depends

on its distance from one end, and on the instant in time.

Its displacementuat any timetcan be expressed as

u=f(x,t),wherexis its distance from 0.

The equation of motion is as stated in Section 53.4 (a),

i.e.

∂^2 u

∂x^2

=

1

c^2

∂^2 u

∂t^2

.

The boundary and initial conditions are:

(i) The string is fixed at both ends, i.e.x=0and

x=Lfor all values of timet.

Hence,u(x,t)becomes:

u( 0 ,t)= 0

u(L,t)= 0

}

for all values oft≥ 0

(ii) If the initial deflection ofPatt=0 is denoted by

f(x)thenu(x, 0 )=f(x)

(iii) Let the initial velocity ofPbeg(x),then

[

∂u

∂t

]

t= 0

=g(x)

Initiallyatrial solutionof the formu(x,t)=X(x)T(t)

is assumed, whereX(x)is a functionofxonly andT(t)

is a function oftonly. The trial solution may be simpli-

fied tou=XTand the variables separated as explained

in the previous section to give:

X′′

X

=

1

c^2

T′′

T

When both sides are equated to a constantμthis results

in two ordinary differential equations:

T′′−c^2 μT=0andX′′−μX= 0

Three cases are possible, depending on the value

ofμ.

Case 1:μ> 0

For convenience, letμ=p^2 ,wherepis a real constant.

Then the equations

X′′−p^2 X=0andT′′−c^2 p^2 T= 0