Angular momentum (sometimes called moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important physical quantity because it is a conserved quantity – the total angular momentum of a closed system remains constant.
The central potential is given by $U(r)=\frac{1}{2} k r^2$, which represents a 3D harmonic oscillator. The Lagrangian for this system in spherical coordinates is:
$$ L=\frac{1}{2} m\left(\dot{r}^2+r^2 \dot{\theta}^2+r^2 \sin ^2 \theta \dot{\phi}^2\right)-\frac{1}{2} k r^2 $$
Since the potential is central, angular momentum is conserved. Let $\lambda=m r^2 \dot{\phi}$ be the magnitude of the angular momentum. The energy is given by:
$$ E=\frac{1}{2} m\left(\dot{r}^2+r^2 \dot{\theta}^2+r^2 \sin ^2 \theta \dot{\phi}^2\right)+\frac{1}{2} k r^2 $$
For simplicity, let's consider motion in the equatorial plane $(\theta=\pi / 2, \dot{\theta}=0)$. Then:
$$ E=\frac{1}{2} m\left(\dot{r}^2+r^2 \dot{\phi}^2\right)+\frac{1}{2} k r^2 $$
Substituting $\dot{\phi}=\lambda /\left(m r^2\right)$, we get:
$$ E=\frac{1}{2} m \dot{r}^2+\frac{\lambda^2}{2 m r^2}+\frac{1}{2} k r^2 $$
Solving for $\dot{r}$ :
$$ \dot{r}=\sqrt{\frac{2}{m}\left(E-\frac{\lambda^2}{2 m r^2}-\frac{1}{2} k r^2\right)} $$
This is consistent with the given eqn. (1).
Now, let's solve this differential equation.
$$ \begin{gathered} \frac{d r}{d t}=\sqrt{\frac{2}{m}\left(E-\frac{\lambda^2}{2 m r^2}-\frac{1}{2} k r^2\right)} \\ \frac{d r}{\sqrt{\frac{2}{m}\left(E-\frac{\lambda^2}{2 m r^2}-\frac{1}{2} k r^2\right)}}=d t \end{gathered} $$
Let $\omega=\sqrt{k / m}$. Then $k=m \omega^2$.
$$ \begin{gathered} E=\frac{1}{2} m \dot{r}^2+\frac{\lambda^2}{2 m r^2}+\frac{1}{2} m \omega^2 r^2 \\ \dot{r}^2=\frac{2 E}{m}-\frac{\lambda^2}{m^2 r^2}-\omega^2 r^2 \end{gathered} $$
Let $r^2=u$. Then $2 r \dot{r}=\dot{u}$, and $\dot{r}=\dot{u} /(2 \sqrt{u})$.
$$ \begin{gathered} \frac{\dot{u}^2}{4 u}=\frac{2 E}{m}-\frac{\lambda^2}{m^2 u}-\omega^2 u \\ \dot{u}^2=4 u\left(\frac{2 E}{m}-\frac{\lambda^2}{m^2 u}-\omega^2 u\right)=\frac{8 E u}{m}-\frac{4 \lambda^2}{m^2}-4 \omega^2 u^2 \\ \dot{u}^2+4 \omega^2 u^2-\frac{8 E}{m} u+\frac{4 \lambda^2}{m^2}=0 \end{gathered} $$
Completing the square: