Finding de-Sitter Schwarzschild Metric

Your metric is:

$$d s^{2}=-e^{2 Phi(r)} d t^{2}+e^{2 Lambda(r)} d r^{2}+r^{2} d Omega^{2}$$

and Einstein's Equation:

$$G_{μν} = -frac{3}{L^{2}}g_{μν}$$

(Assuming that $L$ is the dS radius then the equation represents an AdS spacetime not dS. For a dS spacetime $G_{μν} = frac{3}{L^{2}}g_{μν}$). Clearly there are two unknown functions and one expects two independent equations(the $θθ$ equation is related to $rr$ and $tt$). The $tt,rr,θθ$ equations are the following:

$$frac{e^{-2 (Lambda (r)+Phi (r))} left(-left(L^2+3 r^2right) e^{2 Lambda (r)}-2 L^2 r Lambda '(r)+L^2right)}{L^2 r^2} =0$$

$$frac{e^{-4 Lambda (r)} left(left(L^2+3 r^2right) e^{2 Lambda (r)}-L^2 left(2 r Phi '(r)+1right)right)}{L^2 r^2}=0$$

$$frac{frac{3 r}{L^2}+e^{-2 Lambda (r)} left(left(r Phi '(r)+1right) left(Lambda '(r)-Phi '(r)right)-r Phi ''(r)right)}{r^3} =0$$

The first equation is a differnetial equation for $Λ$. Integrating we get:

$$Lambda (r)=-frac{1}{2} ln left(frac{-c_1+L^2 r+r^3}{L^2 r}right)$$

Using the above the second equation is solved:

$$Phi (r)=frac{1}{2} left(ln left(-c_1+L^2 r+r^3right)-ln (r)right)+c_2$$

,where $c_1,c_2$ are constants of integration and plugging the results in the third equation we can see that $Λ,Φ$ satisfy the equation. The metric will read:

$$d s^{2}=-(frac{e^{2 c_2} left(-c_1+L^2 r+r^3right)}{r}) d t^{2}+(frac{L^2 r}{-c_1+L^2 r+r^3})d r^{2}+r^{2} d Omega^{2}$$

For the appropriate asymptotic behavior we shall identify:

$$c_2 = - ln(L^2)/2$$

and comparing with the vaccum solution: $c_1 = ML^2$.

Now the line element takes the form:

$$ds^{2} =- (1-M/r +frac{r^{2}}{L^2})dt^2 + (1-M/r +frac{r^{2}}{L^2})^{-1}dr^2 + r^2dOmega^2$$

If you never tried to obtain (doing all calculations by hand) at least one of the famous solutions to Einstein's equations i strongly recommend you to do it.