Question regarding Solutions to the equation $y'' + y = y^3$ A while ago I had a question given to me by a tutoring student that read as follows:

A solution to the differential equation $y'' + y = y^3$ is:
A) $y = \tanh(x/\sqrt{2})$
B) $y = \tanh(x\sqrt{2})$
C) $y = \coth(x/\sqrt{2})$
D) $y = \coth(x\sqrt{2})$
E) None of these.
[3 Marks]

Given the low marks available, I presumed that the desired method would be to simply 'plug-in' the given solutions to see which were satisfactory.
Interestingly, I found that both A) and C) satisfied this equation.
I was curious about this, since the formatting to me implied that only one of these should be correct...
I decided to scout around online a bit to see if there was an actual method to solving a DE like this to see if I could narrow it down, or pull these solutions out directly (for my own interest).
The results I obtained this way were interesting. My method is as below
$$y''+y=y^3$$
$$y''y'+yy'=y^3y'$$
$$\frac{1}{2}((y')^2)'+\frac{1}{2}(y^2)'=\frac{1}{4}(y^4)'$$
$$(y')^2+y^2=\frac{1}{2}y^4$$
$$y'=\frac{1}{\sqrt{2}}\sqrt{y^4-2y^2}$$
Separating variables and integrating both sides:
$$\int \frac{dy}{y\sqrt{y^2-2}} = \frac{x}{\sqrt{2}} + C$$
Answers I got from here then vary slightly;
in my initial attempt I made the substitution $y = \sqrt{2}\cosh{u} \rightarrow dy=\sqrt{2}\sinh{u}du$, leading to:
$$\frac{1}{\sqrt{2}}\int{\frac{du}{\cosh{u}}} = \frac{x}{\sqrt{2}}+C$$
Using a further substitution of $v = e^u$ I finally arrived at a solution:
$$\frac{2}{\sqrt{2}}\arctan{e^{\cosh^{-1}{\frac{y}{\sqrt{2}}}}} = \frac{x}{\sqrt{2}}+C$$
After doing some rearranging, I arrived at:
$$y = \sqrt{2}\cosh{\left(\ln\left({\tan{\left(\frac{x}{2}+K_1\right)}}\right)\right)}$$
With $K_1 = C/\sqrt{2}$
Sticking this into wolfram alpha seems to show that this is another satisfactory solution to my original equation.
Further; when I checked an integral calculator for the result of
$$\int\frac{dy}{y\sqrt{y^2-2}}$$
I instead managed to acquire:
$$\frac{1}{\sqrt{2}}\arctan{\left(\frac{1}{\sqrt{2}}\sqrt{y^2-2}\right)}=\frac{x}{\sqrt{2}}+C$$
leading to
$$y = \sqrt{2}\sec(x+K_2)$$
where $K_2 = C\sqrt{2}$
Again, putting this into wolfram alpha seems to indicate that it is too another solution to the equation.

What I'm wondering is



*

*How to extract the two solutions indicated in the question out of the DE, as I've clearly been unable to do so,





*What's the deal with the solutions I've found? I wouldn't be as surprised or intrigued if they both simplified to one (or both) of the results expected from the question, but plotting them indicates that they are not equivalent.


Would be greatful to hear some insight from the wider community.
Thanks!

The first integral is
$$ (y')^2+y^2=\frac12y^4+C_1. \tag1$$
Suppose $y\neq0$. Let $u=y^{-1}$. Then
$$ u'=-y^{-2}y', u''=2y^{-3}(y')^2-y^{-2}y'' $$
and hence
\begin{eqnarray}
u''+u-u^3&=&2y^{-3}(y')^2-y^{-2}y''+y^{-1}-y^{-3}\\
&=&2y^{-3}[-y^2+\frac12y^4+C_1]-y^{-2}(-y+y^3)+y^{-1}-y^{-3}\\
&=&(2C_1-1)y^{-3}.
\end{eqnarray}
Thus for $C_1=\frac12$, then if $y$ is a solution, so is $u=y^{-1}$. Clearly $y=\tanh(x/\sqrt{2})$ satisfies (1) for $C_1=\frac12$.
