Transverse displacement profile of the string using Fourier series

A string of length $$L$$ fixed at $$x=0$$ and $$x=L$$ and released at time time $$t=0$$, the transverse displacement at a position $$x$$ along the string is given by: $$y(x,0)=Ax(L-x)$$

Assuming that the string is released from rest, write down an expression for $$y(x, t)$$, the transverse displacement profile of the string at an arbitrary later time.

My solution:

The initial displacement $$y(x, 0)$$ is can be represented as a sum of harmonic $$y(x,0)=∑_{(n=1)}^∞a_n y_n (x)$$ where $$y_n(x)=\sin⁡(nπx/L)$$ and $$a_n=\frac{4AL^2 (1-(-1)^n )}{(π^3 n^3 )}$$.

This is the part i am struggling with:

$$y(x,t)=∑_{(n=1)}^∞(\frac{4AL^2 (1-(-1)^n )}{(π^3 n^3 )})(\sin⁡(nπx/L)) \cos(πvnt/L)$$

• What exactly are you struggling with? You have a term for the solution, do you believe it to be incorrect? Why don't you plot it for $t=0$ and compare it with $x(L-x)$? Sep 30, 2020 at 18:23

I'm not sure what the problem is, exactly, so let me know if I've got it wrong. I have below a plot for the case $$L=1$$.
1. The blue line represents the function $$y(x) = x(1-x)$$,
2. The orange line represents the function $$y_1(x) = 4 \left(\frac{1- (-1)^1}{\pi^3 1^3}\right) \sin{(\pi x)},$$
3. The green line represent the function $$y_3(x) = \sum_{n=1}^3 4 \left(\frac{1- (-1)^n}{\pi^3 n^3}\right) \sin{(n \pi x)}$$
(I've leave it to you to figure out why I didn't plot $$y_2(x)$$.)