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The given equation is y_tt = 4 y_xx

0 < x < pi, t>0

where y_tt is the 2nd derivative with respect to t, y_xx is 2nd wrt x

Boundary conditions

y(0,t) = 0 and y(pi,t) = 0

And initial conditions

y_t (x,0) = 0 = g(x)

y(x,0) = sin^2 x = f(x)

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General solution (d'Alembert's solution):

y(x,t) = 1/2[F(x+at) - F(x-at)] + int[G(s)ds] from x-at to x+at

My teacher wrote that F(x) is the odd periodic extension of f(x), and then wrote

F(x) = sign(sinx)sin^2 x

Why?