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Maga0, m,na=m / 2n/ 2∑ ∑i= 0 j=0ai,jx2i+ 2 j1xm−2i2xn−2j3( m − 2i+ 2)( m − 2i+ 1)− a(2i+ 2 j)(2i+ 2 j − 1)a 0 = 1;,0( n − 2 j + 2)( n − 2 j + 1)(2i+ 2 j)(2i+ 2 j − 1)i , j = − i−1,ji,j−1(23).Mag1, m,na 0 = 1;a,0i,ja=m / 2n/ 2∑ ∑i= 0 j=0ai,jx2i+2 j+11xm−2i2xn−2j3( m − 2i+ 2)( m − 2i+ 1)− a(2i+ 2 j)(2i+ 2 j + 1)= − i−1,ji,j−1( n − 2 j + 2)( n − 2 j + 1)(2i+ 2 j)(2i+ 2 j + 1)If it is necessary to solve the more general equation ofthe form:∂ 2 2 2u u u+ b∂ + d∂ = 0,∂x ∂x ∂ x2 2 21 2 3where b,d the arbitrary constants. Then recurrentexpressions will be the following:Mag0, m,n=m / 2n/ 2∑ ∑i= 0 j=0ai,jx2i+2 j1xm−2i2xn−2j3a 0 ,0 = 1;( m − 2i+ 2)( m − 2i+ 1)ai,j = −ai−1,jb − ai,j−(2i+ 2 j)(2i+ 2 j − 1)Mag1, m,n=m / 2n/ 2∑ ∑i= 0 j=0ai,jx2i+2 j+11xm−2i2x1n−2j3a 0 ,0 = 1;( m − 2i+ 2)( m − 2i+ 1)ai,j = −ai−1,jb − ai,j−(2i+ 2 j)(2i+ 2 j + 1)1( n − 2 j + 2)( n − 2 j + 1)d(2i+ 2 j)(2i+ 2 j − 1)( n − 2 j + 2)( n − 2 j + 1)d(2i+ 2 j)(2i+ 2 j + 1)It is possible to read on the particular and fundamentalsolutions of equation (20) in (Полянин 2001).However, there are indicated only some separatesolutions, which are similar to those given in thearticle:u = Ax + Bx + Cx + D,1 2 3u = Ax + Bx − ( A + B)x + Cx x + Dx x + Ex x2 2 21 2 3 1 2 1 3 2 3where A, B, C, D,E the arbitrary constants.Annual Proceedings of Vidzeme University College “ICTE in Regional Development”, 2006142