Problem source

A particle moves along the $x$-axis in such a way that its acceleration is $kx \dot{x}\,$ where  $k$ is a positive constant.
When $t = 0$,  $x = d$ (where $d>0$) and $\dot{x} =U\,$.   
\begin{questionparts}   
\item Find $x$ as a function of $t$ in the case $U = kd^2$ and show that $x$ tends to infinity as $t$ tends to $\displaystyle \frac{\pi }{2 dk}\,$.   
\item If $U < 0$, find $x$ as a function of $t$ and show that it tends to a limit, which you should state in terms of $d$ and $U\,$, as $t$ tends to infinity.   
\end{questionparts}

Solution source

\begin{questionparts}

\item $\,$ \begin{align*}
&& \ddot{x} &= kx \dot{x} \\
\Rightarrow && \frac{\d v}{\d x} \dot{x} &= k x \dot{x} \\
\Rightarrow && \int \d v &= \int k x \d x \\
\Rightarrow && v &= \frac12kx^2 + C \\
t=0, x = d, \dot{x} = kd^2: && kd^2 &= \frac12kd^2 + C \\
\Rightarrow && \dot{x} &= \frac12k(x^2+d^2) \\
\Rightarrow && \frac{\d x}{\d t} &= \frac12k(x^2+d^2)  \\
\Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2+d^2)} \d x \\
&&&= \frac{2}{kd}\tan^{-1} \frac{x}{d} \\
\Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d}  + C' \\
t = 0, x = d: && 0 &= \frac{\pi}{2kd} + C' \\
\Rightarrow && t &=  \frac{2}{kd}\tan^{-1} \frac{x}{d}-\frac{\pi}{2kd} 
\end{align*}

As $x \to \infty$, $t \to \frac{2}{kd} \frac{\pi}{2} - \frac{\pi}{2kd}  = \frac{\pi}{2kd} $

\item $\,$ \begin{align*}
&& v &= \frac12kx^2 + C \\
t=0, x = d, \dot{x} = U && U &= \frac12kd^2 + C \\
\Rightarrow && \dot{x} &= \frac12k(x^2-d^2)+U \\
\Rightarrow && \frac{\d x}{\d t} &=\frac12k(x^2-d^2)+U \\
\Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2-d^2)+U} \d x \\
&& &=\frac{2}{k} \int \frac{1}{x^2-d^2+\frac{2U}k} \d x \\
&&&= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \\
\Rightarrow && t &=  \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}}  + C'' \\
t = 0, \dot{x} = d: && 0 &=  \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-d}{d+\sqrt{d^2-\frac{2U}k}}  + C'' \\
\Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \left ( \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \frac{d+\sqrt{d^2-\frac{2U}k}}{ \sqrt{d^2-\frac{2U}k}-d} \right )
\end{align*}

as $t \to \infty$ the denominator needs to head to $0$, ie $x \to -\sqrt{d^2-\frac{2U}k}$

\end{questionparts}