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Ge(n,−wx(17).2(2n− 1)! ∞ ( i + n − 1)! i) = ∑ × w x( n − 1)! (2i+ 2n− 1)! i!i=0THE METHODS COMPARISON2iThe author is confident, that experiencedmathematicians will say - why all these complexitiesand fabrications?Indeed there are excellent methods of integraltransforms, the method of the solutions through Green'sfunction or through the fundamental solution, andfinally the method of separation of variables. Forexample in (Полянин 2001) the following widelyknown particular solutions are given:u = Ax + B C y + D y w =2( )( cos µ sin µ ), µ ,u = Ax + B C hµ y + D hµ y w = −µ2( )( c s ), ,u = A µ x + B µ x Cx + D w = µ2( cos sin )( ), ,u = A hµ x + B hµ x Cx + D w = −µ2( c s )( ), ,u = ( Acos µ x + B sin µ x)( C cos µ y + D sin µ y), w = µ + µ ,2 21 1 2 2 1 2u A x B x C h y D h y w= ( cos µ1+ sin µ1)( c µ2+ s µ2), = µ2 2 1− µ2u = ( Ac hµ x + B s hµ x)( C cos µ y + Dsin µ y), w = − µ + µ ,2 21 1 2 2 1 2u = ( Ac hµ x + B s hµ x)( C c hµ y + D s hµ y), w = −µ − µ ,2 21 1 2 2 1 2,where A, B, C,D - the arbitrary constants. Whyto bother yourself? What is the difference? However,the difference is very essential and fundamental. Itmeans that in the solution which was done by themethod of dual substitution the functional"responsibilities" of the separate function groups aresorted out on the growth of the smallest summarydegree in the total solution of differential equation (as apuff-pastry pie), and in the solutions which are given at(Полянин 2001) all the functions begin from the zero,first or second summary degree of the independentvariables. Let us discuss another solution of theequation of the similar to the equation (10), but in themore common form:∂ u∂x∂ u∂y2 2+ b + w u = 02 2b w – the arbitrary constants. The situation isthe same as it was described above, if the coefficientb is less than zero, then there is an equation ofwhere ,hyperbolic type, and if the coefficient b is more thanzero, then there is an equation of elliptical type. Ifb < 0, w> 0 the equation is already theequation of Klein-Gordon, and not Helmholtz'sequation. The formula (13) will be used to solve thisequation, and functions entering this formula arecalculated from the following recurrent expressions:k / 2Mag0 , ( x,y)= ∑ a Ge(m,wx )a0km=0mk( k − 2m+ 2)( k − 2m+ 1) x= y ; am= −am−1×2m(2m− 1) yk / 2Mag1 , ( x,y)= ∑ a Ge(m + 1, wx )akm = 0m( k − 2m+ 2)( k − 2m+ 1) xx; am= −am1×(2m+ 1)2myk0 = y−Let us examine examples of the solutions ofHelmholtz's equations.222222bbAnnual Proceedings of Vidzeme University College “ICTE in Regional Development”, 2006138