Second order differential equation exercise. Let $y$ be a solution of the equation
$$
y''(t)=y(t)-y(t)^3\,.
$$
Suppose that $y\in L^2(\mathbb R)$ e $y'\in L^2(\mathbb R)$. 
1) Prove that $|y(t)| \leq \sqrt{2}$ for every $t\in \mathbb R$.
2) Prove that either $y(t)=0$ for every $t\in \mathbb R$ or $y(t)$ has constant sign.
 A: By multiplying both terms by $y'$ and integrating from $0$ to $x$ we get:
$$y'^2 = C + y^2 - \frac{1}{2}y^4 \tag{1}$$
as you stated in the comments. If $C\neq 0$ and $y(0)\neq 0$, a qualitative study of $(1)$ gives that the solutions approach the line $y=y_0$, with $\frac{1}{2}y_0^4-y_0^2 = C$. In order to have $f\in L^2$, $C$ must be zero.
Given that:
$$ g(t)=\int\frac{dt}{\sqrt{t^2-t^4/2}}=\log\left(\frac{t}{2+\sqrt{4-2t^2}}\right)$$
we have
$$ g(y(x)) = x+c $$
hence:
$$\frac{y}{2+\sqrt{4-2y^2}}=K e^{x}\tag{2}$$
with $K=\frac{y(0)}{2+\sqrt{4-2y(0)^2}}$. From $(2)$ it follows that $|y|\leq\sqrt{2}$ and :
$$ y(x) = \frac{4K e^x}{1+2K^2 e^{2x}},\tag{3}$$
hence the sign of $y(x)$ is always the same as the sign of $y(0)$, and by assuming $y(0)>0$, $y(x)$ reaches its maximum, $\sqrt{2}$, in $x=-\log(x\sqrt{2})$. Here there are the solutions for $y(0)=\frac{1}{4},\frac{1}{8},\frac{1}{16}$:
$\hspace1in$
They are just translates, as it is obvious from $(1)$. We have:
$$ \| y\|_{L^1} = \pi\sqrt{2}, \quad \|y\|_{L^2}=4,\quad \|y'\|_{L^1}=2\sqrt{2},\quad \|y'\|_{L^2} = \frac{4}{3}.$$
