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I was trying to solve the ODE

\begin{equation} \ddot{r} r = \alpha(\dot{r}^2-1) \end{equation} where $\alpha$ is an arbitrary constant. There are some simple cases when $\alpha = -1 $ then you can use separation of variables to find the solution. For the initial condition when $r'(0)=1$ it also simplifies, $r(t)= t + r(0)$. Not sure how take on the general case. Any help is welcome!

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2 Answers 2

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HINT:

\begin{equation} \ddot{r} r = \alpha(\dot{r}^2-1) \end{equation}

$$\frac{\dot{r}\ddot{r}}{\dot{r}^2-1}=\alpha\frac{\dot{r}}{r}$$

$$\frac{\dot{r}\ddot{r}}{(\dot{r}+1)(\dot{r}-1)}=\alpha\frac{\dot{r}}{r}$$

$$\frac{\ddot{r}}{\dot{r}+1} +\frac{\ddot{r}}{\dot{r}-1}=2\alpha\frac{\dot{r}}{r}$$

Then integrate both side

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Case $1$ : $\alpha=0$

Then $\ddot{r}r=0$

$r=0$ or $\ddot{r}=0$

$r=0$ or $r=C_1t+C_2$

$\therefore r=C_1t+C_2$

Case $2$ : $\alpha\neq0$

Then $\ddot{r}r=\alpha(\dot{r}^2-1)$

$r\dfrac{d^2r}{dt^2}=\alpha\left(\left(\dfrac{dr}{dt}\right)^2-1\right)$

Let $u=\dfrac{dr}{dt}$ ,

Then $\dfrac{d^2r}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=u\dfrac{du}{dr}$

$\therefore ru\dfrac{du}{dr}=\alpha(u^2-1)$

$\dfrac{u}{u^2-1}du=\dfrac{\alpha}{r}dr$

$\int\dfrac{u}{u^2-1}du=\int\dfrac{\alpha}{r}dr$

$\dfrac{1}{2}\ln(u^2-1)=\alpha\ln r+c_1$

$\ln\left(\left(\dfrac{dr}{dt}\right)^2-1\right)=2\alpha\ln r+c_2$

$\left(\dfrac{dr}{dt}\right)^2-1=C_1r^{2\alpha}$

$\dfrac{dr}{dt}=\pm\sqrt{C_1r^{2\alpha}+1}$

$dt=\pm\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}$

$\int dt=\pm\int\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}$

$t=\pm\int_k^r\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}+C_2$

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