Page 95 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS USING FOURIER SERIES AND INTEGRALS 85
Equating coefficients of sine and cosine terms

2
2
E 1 (ω − ω )cos(ωτ) = 0 ω = ω n
n
4C (5.63)
2 2
E 2 (ω − ω )sin(ωτ) = sin(ωτ)
n
ω n
Thus
4C
E 1 = 0 E 2 = ω = ω n (5.64)
2
2
ω n (ω − ω )
n
Combining the homogeneous and particular solutions
4C
B 2n−1 = D 1 cos(ω n τ) + D 2 sin(ω n τ) + sin(ωτ) (5.65)
2
2
ω n (ω − ω )
n
The initial conditions at τ = 0 require that
D 1 = 0

4C(ω/ω n ) (5.66)
D 2 =−
2
2
ω n (ω − ω )
n
The solution for B 2n−1 is
4C ω
B 2n−1 = sin(ω n τ) − sin(ωτ) ,ω = ω n (5.67)
ω n (ω − ω ) ω n
2
2
n
The solution is therefore
∞
sin(ω n ξ) ω
y(ξ, τ) = 4C sin(ω n τ) − sin(ωτ) (5.68)
2
2
ω n (ω − ω ) ω n
n=1 n
When ω = ω n the above is not valid. The form of the particular solution should be chosen as
B P = E 1 τ cos(ωτ) + E 2 τ sin(ωτ) (5.69)

Differentiating and inserting into the differential equation for B 2n−1
4C
2 2 2 2
[E 1 τω + 2E 2 ω n − E 1 τω ]cos(ω n τ) + [E 2 τω − E 2 τω − 2E 1 ω n ]sin(ω n τ) = sin(ω n τ)
n
n
n
n
ω n
(5.70)
Thus
4C
E 2 = 0 E 1 =− (5.71)
2ω 2
n