# Periodic solution of $\dot{x} = a(t) x + b(t)$ with $a$ and $b$ periodic

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 $\exp\left[\int_0^T a(s)\, ds\right] \neq 1$ ?

• 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. – The Count Mar 22 '17 at 0:12
• i can not write here ,, it is difficult – math math Mar 22 '17 at 0:14
• Well, if you can't be bothered, why should anyone else? – The Count Mar 22 '17 at 0:15
• More explanations are provided here – EditPiAf Mar 22 '17 at 18:22

TRICK: First, set the integrating factor to be $$m(t)=\text{exp}\bigg[ \int a(t)dt\bigg]$$ 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, $$x(t)=\text{exp}\bigg[- \int a(t)dt\bigg]\int m(t) b(t)dt$$ 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.
• To test for periodicity, you will need the integration constant, as you need to compare the initial value $x(0)$ to the value $x(T)$. – Lutz Lehmann Mar 22 '17 at 8:52