Show that the equation (dx)/(dt)+x=f(t) , where |f(t)|<M for…
Show that the equation $\dfrac{dx}{dt}+x=f(t)$, where $|f(t)| Solve the linear equation with integrating factor $e^t$; requiring boundedness as $t\to-\infty$ forces the homogeneous constant to vanish, leaving a convolution integral. Periodicity then follows by a change of variable in the integral. With $\mu(t)=e^t$: $\dfrac{d}{dt}(e^tx)=e^tf(t)$, so $x(t)=Ce^{-t}+\displaystyle\int_{-\infty}^t e^{-(t-s)}f(s)\,ds$ (the integral converges since $|f| Since $e^{-t}\to\infty$ as $t\to-\infty$, boundedness on all of $\mathbb{R}$ forces $C=0$. This gives the unique bounded solution \[x(t)=\int_{-\infty}^t e^{-(t-s)}f(s)\,ds,\qquad |x(t)|\le M\ \text{for all }t.\] If $f(t+T)=f(t)$, substituting $u=s-T$ gives $x(t+T)=\displaystyle\int_{-\infty}^t e^{-(t-u)}f(u+T)\,du=\int_{-\infty}^t e^{-(t-u)}f(u)\,du=x(t)$. Final answer: \[\boxed{\,x(t)=\int_{-\infty}^t e^{-(t-s)}f(s)\,ds}\] is the unique bounded solution on $(-\infty,\infty)$; it inherits the period $T$ of $f(t)$ whenever $f$ is $T$-periodic.Explanation of Method
Step 1: General solution
Step 2: Force boundedness as $t o-\infty$
Step 3: Periodicity
Result