Non linear differential equation solve for time t

Question

Hi I have following problem, i have a flow vector field given by:

$$ \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix}y^3 - 4x \\ y^3 - y - 3x \end{pmatrix} $$

and I have to answer the question: Show that $|x(t) - y(t)| \rightarrow 0$ as $t\rightarrow \infty$.

My initial idea is to try to solve the DGL but to be frank I have no idea how.

Answer 1

$$x'=y^{3}-4x$$ $$y'=y^{3}-y-3x$$ Subtract the second from the first $$(x-y)'=-(x-y)$$ Thus $$(x-y)=Ce^{-t}$$ Thus $$\lim_{t\rightarrow\infty}|x-y|=\lim_{t\rightarrow\infty}|C|e^{-t}={0}$$

Answer 2

Assuming you mean $y^3$ instead of $y3$, you can simply observe that

$$\dot x - \dot y = y-x \implies \frac{d}{dt}(x-y)=-(x-y) \implies x-y=e^{-t}$$

and the result follows

Answer 3

Note that with $$ u= x-y$$ you get $$\dot u = -u $$

Thus $$u=u_0 e^{-t} \to 0 \text { as } t\to \infty$$

$$u\to 0 \implies |u|= |x-y|\to 0 $$