= Solution
Start from the constant-coefficient <mean-field dynamo> equation
$$
\partial_t\overline B=\nabla\times(U\times\overline B+\alpha\overline B)+\eta\nabla^2\overline B.
$$
Here $\eta$ may include turbulent as well as microscopic diffusion. Use a local Cartesian model, neglect meridional circulation and curvature, and assume independence of $y$. Represent the <poloidal magnetic field> by a vector potential along $y$ and the <toroidal magnetic field> by its $y$ component:
$$
\overline B=\nabla\times(A\hat y)+B\hat y=(-A_z,B,A_x),\qquad U=\Omega z\hat y.
$$
The shear contributes $(\overline B\cdot\nabla)U=\Omega A_x\hat y$. The poloidal part of $\alpha\nabla\times(B\hat y)$ gives $A_t=\alpha B+\eta\nabla^2A$, while the toroidal part of $\alpha\nabla\times\overline B_{\rm pol}$ is $-\alpha\nabla^2A$. Therefore
$$
B_t=\Omega A_x-\alpha\nabla^2A+\eta\nabla^2B.
$$
Retaining one common transverse dependence $e^{i\ell z}$ changes $\nabla^2$ to $\partial_x^2-\ell^2$, producing the two requested equations. The first coupling is the <alpha effect> regenerating poloidal field from toroidal field; the shear is differential-rotation winding in an <Alpha-Omega dynamo>. Retaining the alpha contribution in the second equation also allows <alpha-squared dynamo> feedback, rather than making the pure Parker shear-dominated approximation.
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