Computing an ODE from Landau book I need some help computing the following ODE. 
Consider the ODE 
\begin{equation}
e^{-\lambda}\left(\frac{\lambda^{\prime}}{r}-\frac{1}{r^{2}}\right)+\frac{1}{r^{2}}=0
\end{equation}
where $\lambda(r,t)$
I start with getting rid of the $e^{-\lambda}$ and try to solve by Separation of Variables, but unfortunately I am stuck shortly after that
\begin{gather*}
e^{-\lambda} \left(\frac{\lambda^{\prime}}{r}-\frac{1}{r^{2}}\right) + \frac{1}{r^{2}}=0 \\
\implies \frac{\lambda^{\prime}}{r}-\frac{1}{r^{2}} + e^{\lambda}\frac{1}{r^{2}} =0 \\
\implies \lambda^{\prime} + e^{\lambda} \frac{1}{r} = \frac{1}{r}
\end{gather*}
after this I want to try and get rid of the $\frac{1}{r}$ from the LHS so that I have $\lambda$ on the left and $r$ on the right and then integrate, but I can't seem to find a way. Any help would be greatly appreciated, as well as pointing me in the right direction would be appreciated. 
The solution to this equation is 
\begin{equation}
e^{-\lambda} = 1 + \frac{\text{const.}}{r}
\end{equation}
Will making the RHS of the above ODE change our approach when attempting to solve the equation. For example, consider that the RHS is no longer equivalently zero, rather it is equal to some constant, namely $\frac{8 \pi k}{c^{4}}T^{0}_{0}$, such that we may rewrite the equation as 
\begin{equation}
e^{-\lambda}\left(\frac{\lambda^{\prime}}{r}-\frac{1}{r^{2}}\right)+\frac{1}{r^{2}}=\frac{8 \pi k}{c^{4}}T^{0}_{0}
\end{equation}
where $k$ is the gravitational constant and as aforementioned $T^{0}_{0}$ is the energy-momentum tensor, nonetheless, this will simply be a number, hence, the RHS may be considered as a constant entirely. Will this modification consent to a change in approach when attempting to solve the equation or may we proceed in a similar fashion, i.e. by Separation of Variables or u-substitution, noting that the $T^{0}_{0}$ is not a function of $r$. 
 A: $$e^{-\lambda}\left(\frac{\lambda^{\prime}}{r}-\frac{1}{r^{2}}\right)+\frac{1}{r^{2}}=0$$
I suppose that $\lambda^{\prime}$ means $\frac{\partial\lambda(r,t)}{\partial r}$ since in the wording of the question you specify that $\lambda$ is function of two variables $r$ and $t$. So "prime" is an ambiguous notation. Nevertheless this notation will continue to be used below insofar $\lambda'=\frac{\partial \lambda(r,t)}{\partial r}$ is accepted.
Change of function :
$$e^{-\lambda}=u\quad;\quad u'=-\lambda'u\quad;\quad \lambda'=-\frac{u'}{u}$$
$$u\left(\frac{-\frac{u'}{u} }{r}-\frac{1}{r^{2}}\right)+\frac{1}{r^{2}}=0$$
$$u'+\frac{u}{r}-\frac{1}{r}=0$$
There is no difficulty to solve this linear ODE :
$$u=1+\frac{c}{r}$$
$$e^{-\lambda}=1+\frac{c}{r}$$
Now, without forgetting that $\lambda$ is also function od $t$ :
$$e^{-\lambda}=1+\frac{f(t)}{r}$$
where $f$ is an arbitrary function (to be determined according to some initial condition).
A: Rewrite your equation
$$
\lambda^{\prime} + e^{\lambda} \frac{1}{r} = \frac{1}{r}
$$
as
\begin{gather*}
\lambda^{\prime}=  \frac{1}{r}\left(1-e^{\lambda}\right).
\end{gather*}
Divide by $\left(1-e^{\lambda}\right)$ and integrate
$$
-\ln(e^{-\lambda}-1)=\ln(r)+K,
$$
where $K$ is a free constant. Solve for $e^{-\lambda}$ by applying the following operations to both sides: multiplying by $-1$, exponentiate, and add 1. You then get
$$
e^{-\lambda}=1+\frac{K'}{r},
$$
where $K'$ is a free constant.
