It is very much like using sinhnyand sinhn1−yis the solution foru1x,y.———————————————————————————————————————Solution to Problem#3Forux,y,z,t, we have∂2ux,y,z,t∂x2∂2ux,y,z,t∂y2∂2ux,y,z,t∂z21v2∂2ux,y,z,t∂t2in the boxed regionRx,y,z|0≤x≤a,0≤y≤b,0≤z≤c,given thatux,y,z,tsatisfies the sixboundary conditionsof the form:u0,y,z,t0,ua,y,z,t0andux,0,z,t0,ux,b,z,t0andux,y,0,t0,ux,y,c,t0and the twoinitial conditions:ux,y,z,0Fx,y,z,∂ux,y,z,t∂tt0Gx,y,z.Now we letux,y,zxyzt, and get′′xyztx′′yztxy′′zt1v2xyz′′tor′′xx′′yy′′zz′′tv2t.This says that′′xxc1,′′yyc2,′′zzc3,′′tc2tc4withc1c2c3c4. Using the boundary condition, these become′′x−c1x0with00anda0and′′y−c2y0with00andb0and′′z−c3z0with00andc0.These lead toc1−ka2andkxsinkxaandc2−mb2andmysinmyband15