Page 409 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes We have now succeeded in separating the variables since the left hand member of (v) is
a function of Y only and hence both members of (v) are equal to a constant which we have
2
called .
2 2
Let 2 then ...(vi)
C
the solutions are
X A 1 sin x B 1 cos x

X A sin x B cos x ...(vii)
2 2
And A s and B s are arbitrary constants. Now, to satisfy the boundary conditions (ii), it is obvious
that there cannot be any cosine forms present so that we must have

B 1 B 2 0
Also we must have

sin a 0

and sin b 0

m
which gives m 0,1,2,.....
a
n
and n 0,1,2,.....
b
From (vi) we find that

m 2 n 2
2
c
mn ...(viii)
a b
Hence for all value of m and n we find a particular solution of (i) that satisfies the boundary
conditions (ii) of the form

m x n y
V B e mnt sin sin
mn
a b
If we sum over all possible values of m and n construct the general solution

m x n y
V B e mnt sin sin ...(ix)
mn
m 1 n 1 a b
Using initial conditions (iii), we get

m x n y
F , x y B mn sin sin ...(x)
m 1 n 1 a b
Multiplying both sides of (x) by

r x s y
sin sin ...(x)
a b

402 LOVELY PROFESSIONAL UNIVERSITY