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ANSWERGiven that{:[P_(tt)-16P_(xx)=0quad-oo < x < oo.quad t > 0],[{:[p(x","0)={[10,|x| <= 1],[0,|x| > 1]:}],[P_(t)(x","0)={[1,|x| <= 1],[0,|x| > 1]}]:}],[" comparing (1) with the following "],[P_(tt)-e^(2)P_(xx)=0quad-oo < x < oo","t > 0", "],[p(x","0)=f(x)quad-oo < x < oo],[" Pt "(x","0)=g(x)","quad-oo < x < oo", "],[" we get "],[c=4quad f(x)={[10,|x| <= 1],[0,|x| > 1]:}],[" and "g(x)={[1,|x| < 1],[0,|x| > 1]:}]:}using D'Atembert's formula the solution at any point is given by{:[P(x","t)=(f(x+ct)+f(x-ct))/(2)],[+(1)/(2c)int_(x-ct)^(x+ct)g(s)ds],[=(f(x+4t)+f(x-4t))/(2)+(1)/(8)int_(-x-4t)^(x+4t)g(s)ds^(p)]:}Now, max{f(x+4t)}=10; which is attained on the strip -1 <= x+4t <= 1.similardy, max {f(x-4t)}=10; ... See the full answer