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Given the first-order linear equation $\dot{x} = a(t) x + b(t)$ where $a$ and $b$ are $T$-periodic functions, show that it has a periodic solution if and only if $\exp\left[\int_0^T a(s)\, ds\right] \neq 1$. Find the periodic solution of the equation $\dot{x} = t x + \sin(t)$.

I solved the homogeneous equation $$ \int_0^T \frac{\dot{x}(s)}{x(s)} \, ds = \ln\left(\frac{x(T)}{x(0)}\right) = \int_0^T a(s) \, ds \, . $$ But how to show $\exp\left[\int_0^T a(s)\, ds\right] \neq 1$ ?

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  • $\begingroup$ Your picture is dark and blurry. You couldn't be bothered to type it in or use a modicum of grammar or mechanics. No, I can't help. $\endgroup$ – The Count Mar 22 '17 at 0:12
  • $\begingroup$ i can not write here ,, it is difficult $\endgroup$ – math math Mar 22 '17 at 0:14
  • $\begingroup$ Well, if you can't be bothered, why should anyone else? $\endgroup$ – The Count Mar 22 '17 at 0:15
  • $\begingroup$ More explanations are provided here $\endgroup$ – EditPiAf Mar 22 '17 at 18:22
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I wouldn't worry about your grammar that can easily be fixed by people with a penchant for it. You can solve the general governing equation by the integrating factor method. If you are confused with this Wolfram Alpha has a lovely page explaining the details on this which helped me during my Bachelors. This is the general method when given a non-homogeneous first order ODE.

TRICK: First, set the integrating factor to be \begin{equation} m(t)=\text{exp}\bigg[ \int a(t)dt\bigg] \end{equation} Recall that this method does not bother with the constant of integration. Then by the product rule and integrating both sides with respect to $t$ will yield the general solution, \begin{equation} x(t)=\text{exp}\bigg[- \int a(t)dt\bigg]\int m(t) b(t)dt \end{equation} HINT: Now, by definition of a periodic solution $x(t)=X(t+T)$ as we have assumed that the solution is periodic in $T$. Note that this definition holds for $a$ and $b$ as well. It should now just be an exercise of plugging in limits when you have got this far.

Hopefully, this helps a little bit it is very late for me and I have tried to direct you down the route I would choose to solve this!

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  • $\begingroup$ can you solve all in detail ? @Buddhaha $\endgroup$ – math math Mar 22 '17 at 3:38
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    $\begingroup$ To test for periodicity, you will need the integration constant, as you need to compare the initial value $x(0)$ to the value $x(T)$. $\endgroup$ – Lutz Lehmann Mar 22 '17 at 8:52
  • $\begingroup$ Can you explain more ? $\endgroup$ – math math Mar 22 '17 at 10:47

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