The $\Lambda$ term

Einstein initially was interested in finding out $\dot{a}=0$ solutions to his equations. This is possible only in a closed universe with $k=+1$ and the matter density appropriately adjusted. But equation $2.17$ prevents $\ddot{a}$ for any Universe with matter with nonnegative pressures to be nonzero. So he introduced a modification to his equation by adding the infamous $\Lambda$ term.

$$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}+\Lambda g_{\mu\nu}=8\pi G T_{\mu\nu}$$ (3.15)

With this introduction we have

$$-3\frac{\ddot{a}}{a}=4\pi G(\rho + 3p)+\Lambda$$ (3.16)

And the equation involving the Hubble constant

$$\left(\frac{\dot{a}}{a}\right)^2=\frac{8}{3}\pi G\rho-\frac{k}{a^{2}}+\frac{\Lambda}{3}$$ (3.17)

This leads to a static Universe even with a nonnegative $\rho$,$p$ and $\Lambda$.The solution is called the static universe solution. But it faces the following problem. The solution requires a great fine tuning of the constants involved and hence any departure from the initial conditions can lead to a departure from the solution. And also in the meantime Hubble announced his discovery that the Universe is expanding and then there was no need of the $\Lambda$ term.