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Differential Equations Lecture Work Solutions 288

# Differential Equations Lecture Work Solutions 288 - 5 4...

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Unformatted text preview: 5 4 x−ct<0 t 3 2 ux (0,t)=h(t) x−ct>0 1 0 u(x,0)=0 and ut (x,0)=0 −1 −5 −4 −3 −2 −1 0 x 1 2 3 4 5 Figure 58: Domain for problem 4 4. utt − c2 uxx = 0 general solution u(x, t) = F (x − ct) + G(x + ct) For x − ct > 0 u(x, t) = 0 since u = ut = 0 on the boundary. For x − ct < 0 we get the inﬂuence of the boundary condition ux (0, t) = h(t) Diﬀerentiate the general solution: ( − ct ux (x, t) = F (x − ct) · 1 + G (x + ct) · 1 = dF(xx− ct)) + d . . . chain rule prime means derivative with respect to argument As x = 0 : h(t) = ux (0, t) = dF (− ct) d(− ct) + dG(ct) d(ct) = −1 c dF (− ct) dt 288 + 1 dG(ct) c dt dG(x + ct) d(x + ct) ...
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