Solve PDE Wave Equation for Stretched String of Length 2

[depthofit]

Homework Statement

Having exams on monday but still having problems with PDE. I thought i got it until i saw the teacher's solution different from mine and i have no idea wtf she doing also.. Contacting her is not an option. Please give me a hand with this.. Much appreciated.

Solve the equation of vibrations of a stretched string of length 2,

d2u/dt2 = 16 * d2u/dx2 (0<x<2)
*2nd order differential

for t>0. The string is clamped at each end, so that the boundary conditions are:

u(0,t) = u(2,t) = 0

At t=0, the initial displacement of the string is,

u(x,0) = 2sin (pi * x)

and its velocity

ut(x,0) = 2sin (pi/2 * x)
*t is a subscript

Homework Equations

The Attempt at a Solution

Below is my attempt,

This is the lecturer's solution,

 

[lippyka]

The PDE can be written in its standard form,du/dt + c2 d2u/dx2 = 0 where c2 = 16The solution of this PDE is of the form u(x,t) = F(x + ct) + G(x - ct)where F and G are functions to be determined.Applying the boundary conditions, u(0,t) = 0 and u(2,t) = 0,gives F(ct) + G(-ct) = 0 which yields F(x) = -G(x).Applying the initial conditions gives,u(x,0) = 2sin (pi * x) = F(x) + G(x)and ut(x,0) = 2sin (pi/2 * x) = c*[F'(x) - G'(x)]Equating the two equations,2sin (pi * x) = F(x) + G(x) and 2sin (pi/2 * x) = c*[F'(x) - G'(x)]Differentiating the first equation,2pi cos (pi * x) = F'(x) + G'(x)Substituting this into the second equation,2sin (pi/2 * x) = c*[F'(x) - G'(x)] = c*[2pi cos (pi * x) - G'(x)]This implies,2sin (pi/2 * x) = 2c pi cos (pi * x) - cG'(x)or G'(x) = 2c pi [cos (pi * x) - sin (pi/2 * x)/c]Integrating w.r.t x,G(x) = 2c pi [sin (pi * x) - (1/2c) cos (pi/2 * x)] + Awhere A is the constant of integration.Therefore, F(x) = -G