Solutions to $1 - \frac{r'^2}{2c^2} + \frac{r r''}{c^2}=0$?

Question

Sage says that$$\sqrt{2 \, c^{2} - e^{K_{1}} r\left(t\right)} e^{\left(-K_{1}\right)} = K_{2} + t$$$$-2\sqrt{2 \, c^{2} - e^{K_{1}} r\left(t\right)} e^{\left(-K_{1}\right)} = K_{2} + t$$are solutions to$$1 - \frac{r'^2}{2c^2} + \frac{r r''}{c^2}=0.$$Is that right?

Answer 1

$$ 1-\frac{r'^2}{2c^2} + \frac{rr''}{c^2} = 0 $$

$$ r'' = r'\dfrac{d}{dr}r' $$ thus $$ 1 - \frac{r'^2}{2c^2} + \frac{r}{c^2}r'\dfrac{d}{dr}r' = \frac{r}{2c^2}\dfrac{d}{dr}r'^2 - \frac{r'^2}{2c^2}+1 $$ let $y = \frac{r'^2}{2c^2}$ we find $$ r\dfrac{dy}{dr} -y+1 = 0 $$ thus $$ \dot{y} -\frac{1}{r}y = -\frac{1}{r} $$ integrating factor methods $$ y\mathrm{e}^{-\ln r} = -\int\frac{1}{r}\mathrm{e}^{-\ln r}dr + C_1 $$ $$ \frac{y}{r} = -\int \frac{1}{r^2}dr + C_1 = \frac{1}{r} +C_1 $$ leads to $$ y = 1 + C_1r = \frac{r'^2}{2c^2} $$ thus $$ r'^2 = 2c^2\left(1 + C_1r\right)\implies r' = \pm\sqrt{1+C_1r}\sqrt{2}c $$ and solve.

Answer 2

After some experimentation, I think I found a way to solve the problem. Start by multiplying both sides by $r'$.

$$r^{-2}r'+\frac1{2c^2}(-r^{-2}r'^3+2r^{-1}r'r'')=0$$

Note that the term in parentheses can be written as

$$(r^{-1}r'^2)'$$

Knowing this, we can integrate both sides to get

$$-r^{-1}+\frac{r^{-1}r'^2}{2c^2}=k_1,\frac{r'^2}{2c^2}=k_1r+1$$ $$r'^2=k_2r+2c^2,r'=\pm\sqrt{k_2r+2c^2}$$ $$\int\frac{dr}{\sqrt{k_2r+2c^2}}=\pm\int dt$$ $$\frac2{k_2}\sqrt{k_2r+2c^2}=\pm t+k_3$$ $$\sqrt{k_2r+2c^2}=\pm\frac{k_2t}2+k_4$$ $$k_2r+2c^2=\frac{k_2^2t^2}4\pm k_2k_4t+k_4^2$$