How do I compute $\displaystyle \frac{d^2t}{d\lambda^2} - \frac{2}{t}\left(\frac{dt}{d\lambda}\right)^2=0.$ can anyone help me on how to compute the following differential equation?
$$\displaystyle \frac{d^2t}{d\lambda^2} - \frac{2}{t}\left(\frac{dt}{d\lambda}\right)^2=0.$$
 A: You have either $t'(λ)=0$ or you can divide to
$$
\frac{t''(λ)}{t'(λ)}-2\frac{t'(λ)}{t(λ)}=0
$$
This is now fully integrable and results in
$$
t'(λ)=Ct(λ)^2,
$$
which is separable and easy to solve.
A: Let $p=\frac{dt}{d\lambda}$, then $\frac{d^2t}{d\lambda^2}=\frac{dp}{d\lambda}=\frac{dp}{dt}\frac{dt}{d\lambda}=\frac{dp}{dt}p$
Now,
$\frac{dp}{dt}p-\frac{2}{t}p^2=p(\frac{dp}{dt}-\frac{2}{t}p)=0$
One solution is $p=0$ which gives $t=c$
The other solution is $\frac{dp}{dt}-\frac{2}{t}p=0$ which gives
$\frac{dp}{p}-\frac{2dt}{t}=0$
$ln\frac{p}{t^2}=c_1$
$\frac{p}{t^2}=c_1$
$\frac{dt}{t^2}=c_1d\lambda$
$\frac{-1}{t}=c_1\lambda+c_2$
Hence, $t=\frac{1}{c_1\lambda+c_2}$ and $t=c$ are solutions.
A: Let $dt/d\lambda=v$. Then notice the useful identity $\frac{dv}{d\lambda}\cdot dt=vdv=\frac{d^2 t}{d\lambda^2}$. Hence, using this to replace $d^2 t/d\lambda^2$, we have $vdv=2v^2 dt/t$. One solution here is $v=dt/d\lambda=0$ (so just constant $t$), but assuming it isn't, then $dv/v=2dt/t$. Integrating should then give $\log v=2\log t+c$, or, $v=ct^2=dt/d\lambda$. Then $cd\lambda=t^{-2}dt$, which integrates to $c_1\lambda+c_2=-1/t$, or, $t=-1/(c_1\lambda+c_2)$. I guess that can be cleaned up to simply $t=\frac{c_1}{\lambda+c_2}$.
