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into (3.7), we get

X k (x)T (t)= X (x)T k (t).
k
Next, we divide both sides of this identity by X k (x)T k (t) and obtain
X (x)

k
=
.
T k (t)
X k (x)
Here we notice that the left-hand side only depends on t, whereas the right-
hand side only depends on x. Hence, both expressions must be equal to a
common constant, i.e.,
T (t) T (t) X (x) k k 3.2 Separation of Variables (3.10)

k = k = −λ k . (3.11)
T k (t) X k (x)
As will become clear below, the minus sign here is introduced for reasons of
convenience. For the time being, we just note that (−λ k ) is some constant
for each pair of functions X k and T k .
From (3.11), we get the following two ordinary differential equations:

X (x)+ λ k X k (x)=0, (3.12)
k

T (t)+ λ k T k (t)=0. (3.13)
k
We first consider (3.12). It follows from the boundary condition (3.8)
that we must have
X k (0) = X k (1)=0. (3.14)
Hence, the functions X k (x) are eigenfunctions of the problem (2.36), with
corresponding eigenvalues λ k . Therefore, from the discussion in Section 2.4
we can conclude that
λ k =(kπ) 2 for k =1, 2,... , (3.15)
and
X k (x) = sin (kπx) for k =1, 2,... . (3.16)
Having solved the second-order problem (3.12), we turn our attention to
the first-order problem (3.13), i.e.,

T (t)+ λ k T k (t)=0.
k
This problem has a solution of the form
2
−(kπ) t
−λ k t
T k (t)= e = e , (3.17)
where we have chosen the constant to be equal to one. Now it follows by
(3.16) and (3.17) that
2
−(kπ) t
u k (x, t)= e sin (kπx) for k =1, 2,... . (3.18)
This is the family {u k } of particular solutions we have been looking for.