Solution for Vibrating String Problem: Wave Equation Problem Explained

[Xyius]

This is the problem, it says to solve the solution to the vibrating string problem.
\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2}
u(0,t)=u(1,t)=0,t>0
u(x,0)=x(1-x),0<x<1
\frac{\partial u}{\partial t}(x,0)=sin(7\pi x),0<x<1

The solution form I obtained (without showing my work.) is..

\sum_{n=1}^\infty [a_ncos(n\pi t)+b_nsin(n\pi t)]sin(n\pi x)

I KNOW this is correct because in the chapter they derive the solution and it is of this form. The answer in the back of the book says this however..

u(x,t)=\frac{1}{7\pi}sin(7\pi t)sin(7\pi x)+\sum_{n=0}^\infty \frac{8}{((2n+1)\pi)^3}cos(2n+1)sin(2n+1)

I know I didn't show my final answer, but it turned out being wrong. I do not understand where they got the first term and most importantly, why "2n+1" appears in the arguments of the sine and cosine terms. That would mean that it would not be following the formula for the wave equation solution, namely..

\sum_{n=1}^\infty [a_ncos(\frac{n \pi \alpha}{L} t)+b_nsin(\frac{n \pi \alpha}{L} t)]sin(\frac{n \pi}{L} x)

Can anyone explain this? Thanks a lot!

 

[Xyius]

Ohh! Never mind! I see what they did. The variable they used was k not n. They replaced n with 2k+1 to eliminate the (-1)^n that was in the solution.

The only reason you can do this I think is because in the solution I had (which I know now is correct) the coefficient formula has a 1-(-1)^n in the numerator. So all even values of n are therefore equal to zero.