The hydrogen atom is easy to solve because it's a one body problem (assuming we take the nucleus as fixed). In that case the eigenfunctions of the Hamiltonian are the usual $1s$, $2s$, $2p$, etc.
When we are considering multielectron atoms we make the approximation that we can write the total wavefunction as a product of hydrogenic atomic orbitals, so for boron we might write:
$$ \Psi_B = \psi_{1s}(e_1)\,\psi_{1s}(e_2)\,\psi_{2s}(e_3)\,\psi_{2s}(e_4)\,\psi_{2p}(e_5) $$
where the $\psi_{1s}$ etc are atomic orbitals that resemble the hydrogen atomic orbitals. But there are two key points you need to note:
this is only an approximation because the repulsion between the five electrons causes the atomic orbitals to mix with each other - there are no distinct atomic orbitals
the atomic orbitals are not the same as the hydrogen atomic orbitals. They are not even a simple scaling of the hydrogen atomic arbitals.
To calculate the wavefunction for e.g. boron we would start by using a Hartree-Fock calculation to get a set of atomic orbitals that give the best approximation to $\Psi_B$. Then we would use a configuration interaction calculation to work out how the HF atomic orbitals mix. The end result is a wavefunction $\Psi_B$ that cannot be factorised into separate atomic orbitals.
So there is no simple equation of the type you describe. For all multielectron atoms the calculation of the wavefunction is a complex process (though straightforward on modern computers).
