+\gamma_{3}^{2}) U_{3}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t )\;+\;i\gamma_{3} P( \gamma_{1}, \gamma_{2}, \gamma_{3}, t )\;+\; \tilde{F}_{3}( \gamma_{1}, \gamma_{2}, \gamma_{3},t )$$

$$\label{eqn137}
\gamma_{1} U_{1}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t ) \;+\; \gamma_{2}\, U_{2}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t ) \;+\; \gamma_{3}\, U_{3}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t ) \;=\;0$$

$$\label{eqn138}
U_{1}(\gamma_{1}, \gamma_{2},  \gamma_{3},  0)\;=\; U_{1}^{0}(\gamma_{1} ,\gamma_{2} ,\gamma_{3})$$

$$\label{eqn139}
U_{2}(\gamma_{1}, \gamma_{2},  \gamma_{3},  0)\;=\; U_{2}^{0}(\gamma_{1} ,\gamma_{2} ,\gamma_{3})$$

$$\label{eqn140}
U_{3}(\gamma_{1}, \gamma_{2},  \gamma_{3},  0)\;=\; U_{3}^{0}(\gamma_{1} ,\gamma_{2} ,\gamma_{3})$$

Hence, we have received a system of linear ordinary differential equations with constant coefficients $(\ref{eqn134})\;-\; (\ref{eqn140})\;$ according to Fourier transforms. At the same time the initial conditions are set only for Fourier transforms of velocity components $U_{1}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t ), \;U_{2}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t ), \;U_{3}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t )$. Because of that we can eliminate Fourier tranform for pressure $P( \gamma_{1}, \gamma_{2}, \gamma_{3}, t )$ from equations $(\ref{eqn134})\;-\; (\ref{eqn136})\;$ on the **second step** of solution. From here assuming that $\gamma_{1} \neq 0,\;\gamma_{2} \neq 0,\;\gamma_{3} \neq 0$, we eliminate $P( \gamma_{1}, \gamma_{2}, \gamma_{3}, t )$ from equations $(\ref{eqn134})\;-\; (\ref{eqn136})\;$ and find:

$$\begin{aligned}
\label{eqn141}
\frac{d}{dt} {\mbox{\large [ \normalsize}} U_{2}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t )\; -\;\frac{\gamma_{2}}{\gamma_{1}} \,U_{1}( \gamma_{1}, \gamma_{2}, \gamma_{3}, t) {\mbox{\large ] \normalsize}} \;= -\nu( \gamma_{1}^{2} +\gamma_{2}^{2} +\gamma_{3}^{2}) {\mbox{\large [ 