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« **on:** February 15, 2018, 09:51:06 PM »
General solution for $ x >-t $: $ u = \psi(x+2t) + \phi(x-2t) $

Given initial conditions, we have $ \psi(x) = \phi(x) = 0 $ so $$ u = 0, x > 2t $$

Using the boundary conditions for $ -t < x < 2t $, ie where $\phi(x), x<0 $

$$ \psi'(t) + \phi'(-3t) = \sin(t) $$

From previous solution $$ \psi(x) = 0 \implies \psi'(x) = 0 $$

So $$ \phi(t) = -\cos(\frac{-t}{3}) + constant, t <0 $$

$$ \phi(t) = -\cos(\frac{t}{3}) + constant, t<0 $$

$$ \phi(x - 2t) = -\cos(\frac{(x-2t)}{3}) + constant $$

For $\phi$ to be constant at x = 2t, ie the same value from both sides of the characteristic equation:

$$ 0 = -\cos(0) + c \implies c = 1 $$

So $ u = -\cos(\frac{(x-2t)}{3}) + 1 , -t <x<2t $