differential equation $y''(x)-ay^3(x)+by(x)=0$ Hi I am trying to find a solution $y(x)$ to this non linear differential equation
$$
y''(x)-ay^3(x)+by(x)=0.
$$
I know a nice solution exists, however how can I go about solving this?  I know non linear ODE's a are tougher than linaer ODE's, however this one has such a closed form solution I thought maybe it would be straight forward.  
I can verify that a solution exists , we can write it in the general form
$$
y(x)=A \tanh cx,\quad y''(x)=2c^2A\tanh cx(\tanh^2 cx-1).
$$
We can simply verify this works by plugging into the ODE to obtain$$
2c^2A\tanh^3 cx-2c^2A\tanh cx-aA^3\tanh^3 cx+bA\tanh cx=0,\quad a,b\in \mathbb{R}.
$$
We can see this is true if 
$$
2c^2 A=aA^3,\quad 2c^2A=bA,\to A=\sqrt\frac{2c^2}{a}=\sqrt \frac{b}{a},\quad c=\sqrt \frac{b}{2}.
$$
We can now see
$
y(x)=A\tanh cx 
$ is a solution for these values of A and c.  
Now how can we solve this differential equation and obtain y(x)?  Thanks for reading this and thinking about it
 A: Maple 18 finds the solution in terms of a Jacobi elliptic function:
$$y \left( x \right) =C_{{2}}\sqrt {2}\sqrt {{\frac {b}{a{C_{{2}}}^{2}-a
+2\,b}}}{\it JacobiSN} \left(  \left( 1/2\,\sqrt {-2\,a+4\,b}x+C_{{1}}
 \right) \sqrt {2}\sqrt {{\frac {b}{a{C_{{2}}}^{2}-a+2\,b}}},{\frac {C
_{{2}}\sqrt {- \left( a-2\,b \right) a}}{a-2\,b}} \right) 
$$
A: Your equation is
$$
y''-ay^3+by=0.
$$
Make a substitution
$$
u(y)=y',
$$
which implies that
$$
y''=u'u.
$$
You get the equation
$$
u'u=ay^3-by,
$$
which integrates to
$$
\frac{u^2}{2}=\frac{ay^4}{4}-\frac{by^2}{2}+C_1,
$$
or
$$
u=\pm\sqrt{\frac{ay^4}{2}-by^2+C_1}\,.
$$
Now you end up with the equation
$$
y'=\pm\sqrt{\frac{ay^4}{2}-by^2+C_1},
$$
which is obviously a separable equation, but the closed form solutions exist only for specific values of $a,b,C_1$.
A: In the Ginzburg-Landau theory, the above equation can be obtained by deriving the Gibbs Free Energy and by setting it equal to zero. The solution can explain the behavior of the order parameter close to the normal-superconductor interface. Since deep inside the superconductor the density of super-electron is constant, one can apply the boundary condition u=0 for x-->inf. Applying this condition to the above equation (the fifth counting from the top) one can easily found that C = 1/2 and the solution can be written in its closed form.
