Solving the 'easy' differential equation $(1 - \phi^2)\phi'' + \phi(\phi')^2 =0$.

Just a lot of pattern matching and manipulation. Rewrite the equation as

$$\frac{\phi''}{\phi'} = -\frac{\phi \, \phi'}{1-\phi^2}$$

This can be written as

$$\frac{d}{dx} \log{\phi'} = \frac12 \frac{d}{dx} \log{(1-\phi^2)}$$

This may be integrated and subsequently exponentiated to produce

$$\phi' = A \left (1-\phi^2\right)^{1/2}$$

where $A$ is a constant of integration. We may then rewrite this equation in differential form as

$$\frac{d\phi}{\left (1-\phi^2\right)^{1/2}} = A \, dx$$

which integrates to

$$\arcsin{\phi} = A x + B$$

where $B$ is another constant of integration. The solution to the above equation is then

$$\phi(x) = \sin{(A x+B)}$$

You may verify that this is indeed the solution by plugging it back into the original equation.