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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.

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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 $\lambda(0) = 0$.

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  • $\begingroup$ how to solve it without knowing general solution or writing the equation of great circles in stereographic projection? $\endgroup$
    – user360777
    Apr 26 '17 at 17:24
  • $\begingroup$ I have added something to get you started $\endgroup$
    – mch56
    Apr 26 '17 at 17:40
  • $\begingroup$ i got it, thanks! in the second equation $\frac{d\mu}{d \lambda}(\lambda^2 +1)\lambda$ should be corrected as $\frac{d\mu}{d \lambda}(\lambda^2 +1)\mu$ $\endgroup$
    – user360777
    Apr 26 '17 at 17:50
  • $\begingroup$ Ah yes. Typo amended and full solution added for you. $\endgroup$
    – mch56
    Apr 26 '17 at 17:51

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