geodesic equations for spheres I get an ordinary differential equation when i'm checking the geodesics of spheres are great circles using stereographic projection (i know there're better ways to get the geodesics directly).
I just wanna know the property of this equation
$\lambda ''(1+\lambda ^2)=\lambda \lambda '^2$
with initial value $\lambda(0)=0$,
such as if it is solvable, and if $\lambda'$ is increasing.
 A: The general solution is $$\lambda = \sinh\left(C_1t+C_2\right)$$
Edit
Since, the OP wants to know how to solve in general.
We have $$(1+\lambda(t)^2)\lambda''(t) = \lambda(t)\lambda'(t)^2$$
Let $\mu(\lambda) = \lambda'(t)$, and hence via the chain rule we can reduce our equation to
$$\frac{d\mu}{d \lambda}(\lambda^2 +1)\lambda = \lambda \mu^2.$$
Hence we have
$$\mu\left(\frac{d\mu}{d \lambda}\lambda^2 + \frac{d\mu}{d \lambda} - \mu \lambda\right) = 0$$
So we either have $\mu= 0$ or $\frac{d\mu}{d \lambda}\lambda^2 + \frac{d\mu}{d \lambda} - \mu \lambda = 0$.
Rearranging the second equation we have can see that
$$\frac{d\mu}{d \lambda} = \frac{\mu \lambda}{\lambda^2+1}$$
is separable. Integrating this gives us
$$\mu =  C_1 \sqrt{\lambda^2 +1}.$$ Thus, all that remains to be done is solve
$$\lambda' = C_1 \sqrt{\lambda^2 +1}$$ which again is seperable and leads to
$$\int \frac{d \lambda}{\sqrt{\lambda^2 +1}} = \int C_1 \, \, dt.$$
This has general solution $\lambda = \sinh(C_1 t + C_2)$. Notice we have two constants as it is a second order differential equation. To solve for one constant, substitute in the condition $\lambda(0) = 0$.
