Higher Engineering Mathematics

516 DIFFERENTIAL EQUATIONS

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 =0 i.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 =0 i.e.m^2 =−4 from which,

m=

√

− 4 =±j 2

Thus, the general solution is:

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

Now try the following exercise.

Exercise 201 Further problems on revising

the solution of ordinary differential equation

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


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


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 proper-
ties, i.e. the string satisfies Hooke’s law. Figure 53.1
shows a flexible elastic string stretched between two
points atx=0 andx=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 fixed. The position of any pointP
on the string depends on its distance from one end,
and on the instant in time. Its displacementuat any

x

0 L x

u(x, t)

u

=

f(x

, t

)

P

Figure 53.1

timetcan be expressed asu=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=0 and

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

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

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

[

∂u

∂t

]

t= 0

=g(x)

Initially atrial solutionof the formu(x,t)=X(x)T(t)

is assumed, whereX(x) is a function ofxonly and

T(t) is a function oftonly. The trial solution may be

simplified 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= 0 and X′′−μX= 0

Three cases are possible, depending on the

value ofμ.

Case 1:μ> 0

For convenience, let μ=p^2 , wherep is a real

constant. Then the equations

X′′−p^2 X= 0 and T′′−c^2 p^2 T= 0

have solutions:X=Aepx+Be−pxand

T=Cecpt+De−cpt where A, B, C and D are

constants.