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(31)The exact remedy for �� = one is u(x, t) = ex?tand v(x, t) = ex+t.To obtain the answer of (thirty) by NHPM, we construct the next homotopy:(1?p)(Dt��U(x,t)?u0(x,t))??+p(Dt��U(x,t)+?2U(x,t)?x2????????2e2tsin(2t)U(x,t)?U(x,t)?x???????+sin(2t)?(U(x,t)V(x,t))?x)=0,(one?p)(Dt��V(x,t)?v0(x,t))??+p(Dt��V(x,t)??2V(x,t)?x2???????+2e?2tcos?(2t)V(x,t)?V(x,t)?x????????cos?(2t)?(U(x,t)V(x,t))?x)=0.(32)Applying selleck compound INNO-406 the inverse operator Jt�� of Dt�� on the two sides from the over equation, we obtainU(x,t)=U(x,0)+Jt��u0(x,t)??pJt��(u0(x,t)+?2U(x,t)?x2????????2e2tsin(2t)U(x,t)?U(x,t)?x???????+sin(2t)?(U(x,t)V(x,t))?x),V(x,t)=V(x,0)+Jt��v0(x,t)??pJt��(v0(x,t)??2V(x,t)?x2???????+2e?2tcos?(2t)V(x,t)?V(x,t)?x????????cos?(2t)?(U(x,t)V(x,t))?x).



(33)For solving system (33), by new homotopy perturbation process, we utilize the Taylor series ofsin(2t)=��n=0��(?1)n(2t)2n+1(2n+1)!,cos?(2t)=��n=0��(?1)n(2t)2n(2n)!,exp?(2t)=��n=0��(2t)nn!,exp?(?2t)=��n=0��(?1)n(2t)nn!.(34)The solution of (30) has the following kind:U(x,t)=U0(x,t)+pU1(x,t),V(x,t)=V0(x,t)+pV1(x,t).(35)Substituting (34) and (35) in (33) and equating the coefficients of like powers of p, we get the following set of equations:U0(x,t)=U(x,0)+Jt��u0(x,t),V0(x,t)=V(x,0)+Jt��v0(x,t),U1(x,t)=Jt��(?u0(x,t)??2U0(x,t)?x2?????+2��n=0��(2t)nn!��n=0��(?one)n(2t)2n+1(2n+1)!U0(x,t)?U0(x,t)?x??????��n=0��(?one)n(2t)2n+1(2n+1)!?U0(x,t)V0(x,t)?x),V1(x,t)=Jt��(?v0(x,t)+?2V0(x,t)?x2??????2��n=0��(?one)n(2t)nn!��n=0��(?one)n(2t)2n(2n)!?????��V0(x,t)?V0(x,t)?x?????+��n=0��(?1)n(2t)2n(2n)!?U0(x,t)V0(x,t)?x).



(36)Assuming u0(x, t) = ��n=0��an(x)pn(t), v0(x, t) = ��n=0��bn(x)pn(t), pn(t) = tn��, U(x, 0) = u(x, 0), and V(x, 0) = v(x, 0) and solving the above equation for U1(x, t) and V1(x, t) bring about theFluconazole resultU1(x,t)=(?a0(x)?ex)t����(��+1)?+(?a1(x)?d2a0(x)dx2)��(��+1)t2����(2��+1)?+(?a2(x)?12d2a1(x)dx2?2b0(x)ex????+2a0(x)ex+2exda0(x)dx?????2ex(db0(x)dx)+8e2x)��(2��+1)t3����(3��+1)?+(?a3(x)?2a0(x)(db0(x)dx)????+?+8a0(x)ex)?����(3��+1)t4����(4��+1)+?,V1(x,t)=(?b0(x)+ex)t����(��+1)?+(?b1(x)?b0(x)ex+a0(x)ex??????+exda0(x)dx?exdb0(x)dx??????+d2b0(x)dx2+4e2x)��(��+1)t2����(2��+1)?+(?b2(x)?12exdb1(x)dx?????12b1(x)ex+12a1(x)ex????+a0(x)db0(x)dx+4b0(x)ex)?����(2��+1)t3����(3��+1)?+(?b1(x)db0(x)dx?13b2(x)ex????+?+13d2b2(x)dx2)��(3��+1)t4����(4��+1)+?.(37)Vanishing U1(x, t) and V1(x, t) lets the coefficients ai, bi, i = 0,one, two,�� have the following values:a0(x)=?ex,??a1(x)=ex,??a2(x)=?12!ex,a3(x)=13!ex,??a4(x)=?14!ex,��,b0(x)=ex,??b1(x)=ex,??b2(x)=12!ex,b3(x)=13!ex,??b4(x)=14!ex,��.