Substituting the results above into (30), we can obtain the following exact solutions to ��(b?1e?��+b0+b1e��)?1,(34)where?+b1(3ak2??3a2k4?3akc)3ake��)??+b0(3ak2??3a2k4?3akc)3ak??��(b?1e?��+b1e��)?1,(33)U2(��)=(b?1(9ak3��k?3a2k4?3akc+3c)6ak2��3?3a2k4?3akce?��?+b1(3ak2??3a2k4?3akc)3ake��)??(28):U1(��)=(?b?1(2ak3��2k?3a2k4?3akc?c)3ak2��?3a2k4?3akce?�� Tipifarnib manufacturer �� = kx + (ct��/��(1 + ��)) + ��0. If we especially take c = ?ak3, b1 = b?1 or c = ?ak3, b1 = ?b?1 in (33), then we obtain the hyperbolic function solitary wave solutions:U3(��)=ktanh��,U4(��)=kcoth?��.(35)Case 2 ��LetU(��)=a?me?m��+a0+amem��b?me?m��+b0+bmem��.(36)Substituting (36) into (28), eliminating the denominator, and collecting all the terms with the same power of epn�� together, equating each coefficient to zero, yield a set of algebraic equations.

Solving these equations yields the following results. Family3. ??a0=0,??b0=0,a?m=?b?m(2am2k3��2mk?3a2m2k4?3akc?c)3amk2��?3a2m2k4?3akc,(37)where?Consideram=bm(3amk2??3a2m2k4?3akc)3ak, bm, b?m are arbitrary constants. Family4. Consideram=bm(3amk2??3a2k4?3akc)3ak,a0=b0(3amk2??3a2m2k4?3akc)3ak,a?m=b?m(9am2k3��mk?3a2m2k4?3akc+3c)6amk2��3?3a2m2k4?3akc,(38)where bm, b0, b?m are arbitrary constants.Substituting the results above into (36), we can obtain the following exact solutions to ��(b?me?m��+b0+bmem��)?1,(40)where??+b0(3amk2??3a2m2k4?3akc)3ak)???+bm(3amk2??3a2k4?3akc)3akem��???=(b?m(9am2k3��mk?3a2m2k4?3akc+3c)6amk2��3?3a2m2k4?3akce?��?��(b?me?m��+bmem��)?1,(39)U6(��)??+bm(3amk2??3a2m2k4?3akc)3akem��)???=(?b?m(2am2k3��2mk?3a2m2k4?3akc?c)3amk2��?3a2m2k4?3akce?m��?(28):U5(��) �� = kx + (ct��/��(1 + ��)) + ��0.

If we especially take c = ?am2k3, bm = b?m or c = ?am2k3, bm = ?b?m in (39), then we obtain the hyperbolic function solitary wave solutions:U7(��)=mktanh��,U8(��)=mkcoth?��.(41)Now we suppose that the solution of (28) can be expressed byU(��)=��p=?m1m2apeip��q=?n1n2bqeiq��.(42)By the balancing process for (28), we have m1 = n1 and m2 = n2. Similar to AV-951 the above, we will also select two cases for computation. Case 1 ��Let m1 = n1 = 1, m2 = n2 = 1. ThenU(��)=a?1e?i��+a0+a1ei��b?1e?i��+b0+b1ei��.(43)Substituting (43) into (28), eliminating the denominator, and collecting all the terms with the same power of eip�� together, equating each coefficient to zero, yield a set of algebraic equations. Solving these equations, we obtain the following two families of values. Family5. ?Considera1=b1(3iak2?3a2k4?3akc)3ak,a0=0,a?1=?b?1(?2ak3��2ik3a2k4?3akc???c)3iak2��3a2k4?3akc,(44)where? b1, b?1 are arbitrary constants. Family6. Considera1=b1(3iak2?3a2k4?3akc)3ak,a0=b0(3iak2?3a2k4?3akc)3ak,a?1=b?1(?9ak3��ik3a2k4?3akc+3c)6iak2��33a2k4?3akc,(45)where b1, b0, b?1 are arbitrary constants.