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Fourier coefficients of the series of the left hand side of (8.45) are equal to the Fourier
coefficients of the series of the right hand side. In particular, for n = 0 we obtain the
ODE:
00
T = 0, (8.46)
0
whose general solution is T 0 (t) = A 0 + B 0 t. Similarly we obtain for n = 2
2
00
T + 4π T 2 = cos 2πt. (8.47)
2
The general solution of this linear nonhomogeneous second-order ODE is

t
T 2 (t) = A 2 cos 2πt + B 2 sin 2πt + sin 2πt.
4π
For n 6= 0, 2, we have
00
2 2
T + n π T n = 0 ∀n 6= 0, 2. (8.48)
n
The solution is T n (t) = A n cos nπt+B n sin nπt for all n 6= 0, 2. Substituting the solutions
of (8.46), (8.47), and (8.48) into (8.44) implies that the solution of the problem is of the
form

∞
A 0 + B 0 t t X
u(x, t) = + sin 2πt cos 2πx+ (A n cos πnt+B n sin πnt) cos πnx. (8.49)
2 4π
n=1
Substituting (8.49) into the first initial condition (8.43), we get
∞
X 1 1
A 0 2
u(x, 0) = + A n cos πnx = cos πx = + cos 2πx,
2 2 2
n=1
therefore,
1
A 0 = 1, A 2 = , A n = 0 ∀n 6= 0, 2.
2
By differentiating (term-by-term) the solution u with respect to t and substituting u t (x, 0)
into the second initial condition of (8.43), we find
∞
X
B 0
u t (x, 0) = + πnB n cos πnx = 2 cos 2πx.
2
n=1
Hence,
1
B 2 = , B n = 0∀n 6= 2.
π
Finally

1 1 t + 4
u(x, t) = + cos 2πt + sin 2πt cos 2πx.
2 2 4π
It is clear that this solution is classical, since the (generalized) Fourier series of the solution
has only a finite number of nonzero smooth terms, and therefore all the formal operations
are justified. Note that the amplitude of the vibrating string grows linearly in t and it is
unbounded as t → ∞.

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