# Given a solution flow to find periodic solutions

Given the system of differential equations $x' = 2x + y^3$ and $y' = -y$ i found the flow $$\phi_t(x,y) = ((x_0 + 1/5y_0^3)e^{2t} - 1/5 y_0^3e^{-3t}, y_0 e^{-t})$$. I am wondering are there any periodic solutions? how do i check?

## 2 Answers

Since $y = y_0e^{-t}$, there are no periodic solutions. If a solution had period $T > 0$, then $y_0^{-(t + T)} = y_0 e^{-t}$, clearly impossible unless $y_0 = 0$. If $y_0 = 0$, $x = x_0 e^{2t}$ and a similar argument applies.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• so if $y_0$ = 0 ...it is periodic solution? – afsdf dfsaf Mar 7 '14 at 4:36
• @afsdf dfsaf: you have to be careful about the meaning and definition of periodic. U sually it means something like there is a smallest $T > 0$ such that $\phi_{t + T}(p) = \phi_t(p)$. This rules out equilibria, since there is no smallest such $T$. If $y_0 = 0$, we still have to deal with $x$. If $x = y = 0$, we have an equilibrium, and no smallest $T > 0$. – Robert Lewis Mar 7 '14 at 5:05
• but if $y_0$ = 0...it has a periodic solution right? – afsdf dfsaf Mar 7 '14 at 5:07
• @afsdf dfsaf: no, because $x$ is not periodic! By the way, how did you arrive at the handle "afsdf dfsaf"? – Robert Lewis Mar 7 '14 at 5:09

The system of differential equations $$x' = 2x + y^3$$ and $$y' = -y$$ As answered by Robert Lewis, the second equation does not make any problem and its solution is $y = y_0e^{-t}$. This first result can be inserted in the first equation, the solution of which being easily obtained using the method of variation of parameters. Its solution is given by $$x= x_0 e^{2 t}+\frac{1}{5} y_0^3 \left(e^{2 t}-e^{-3 t}\right)$$ This gives a parametrized solution of $x(t)$ and $y(t)$. There is no place for any periodic solutions.