Let $ f$ be a periodic function of period $T$ and $\phi $ an solution of the linear equation

$$y'-f(x)y =0$$

Prove that if $ \phi(x)$ is a solution, then $\phi(x+T) $ is also a solution. More than this, prove that there exists a constant $c$ such that

$$\phi(x+T) = c \phi(x)$$

My attempt

Solving the linear ode, we have that $$y=k \exp(\int - f(x) dx)$$

If $\phi$ is a solution, then $\phi(x) =k \exp(\int - f(x) dx)$

I dont know how to rigorously proceed from here: We know that f(x)=f(x+T) since the function is periodic. So, is it sufficient to substitute f(x+T) for f(x) when calculating \phi(x+T)? I am not sure how I should work with periodic functions and indefinitive integrals. Maybe it is more convenient tp write

$$y= k \exp(\int -f(x)dx + C_1)$$

and when calculating $\phi(x+T)$, this constant $C_1$ changes for another constant, we say $C_2.$. But zgain, I am not sure if it is correct.



Suppose $\phi$ is a solution, and let $\psi(x)=\phi(x+T)$. Then, $$\psi'(x)-f(x)\psi(x)=\phi'(x+T)-f(x)\phi(x+T)=\phi'(x+T)-f(x+T)\phi(x+T)=0$$ because $f$ is periodic and $\phi$ is a solution. Hence, $\phi(x+T)$ is a solution.

For the other part, simply substitute the ODE to find that $$\frac{d}{dt}\left(\frac{\phi(x+T)}{\phi(x)}\right)=0.$$

  • $\begingroup$ Thanks for the first part! But what do you mean by substitute the ODE? $\endgroup$ – Giiovanna Nov 8 '14 at 16:03
  • $\begingroup$ Oh, I got it. Thanks for your help $\endgroup$ – Giiovanna Nov 8 '14 at 16:06
  • 1
    $\begingroup$ I mean replace $\phi'(x)$ by $f(x)\phi(x)$ and $\phi'(x+T)$ by $f(x)\phi(x+T)$. $\endgroup$ – Spenser Nov 8 '14 at 16:07

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