$$ \therefore x-x_0 = \pm \int_{\phi(x_0)}^{\phi(x)} \frac{d \phi}{\sqrt\frac{\lambda}{2}\left( \phi^2-(\frac{m}{\sqrt \lambda})^2\right)} $$ How can we write the above equation to as, $$ \phi(x) = \pm \frac{m}{\sqrt \lambda} \tanh\left[\frac{m}{ \sqrt 2} (x-x_0)\right]$$
Do I need to use the hyperbolic function $\tanh$ ? I tried but got different!
Note that: $$\int \frac{dx}{x^2 - a^2} = \frac{\rm{arctanh}(x/a)}{a}$$ Define $a= m / \sqrt{\lambda},\ b = \sqrt{\lambda / 2}$: $$\pm ab(x-x_0) = \rm{arctanh}(\phi(x) / a) - \rm{arctanh}(\phi(x_0) / a)$$ Since we don't know what $\phi(x_0)$ is, call $\rm{arctanh}(\phi(x_0) / a)$, $c$, and get: $$\phi(x) = a \tanh\left(c\pm ab(x-x_0)\right)$$ $$\phi(x) = \frac{m}{\sqrt{\lambda}} \tanh\left(c\pm \frac{m}{\sqrt{2}}(x-x_0)\right)$$ So if you assume that $\phi(x_0)=0$, you can get the desired result.