Dtyjlui

I am revising some exercises from a course on ODE's that I have not managed to do. This exercise asks the reader to show that any solution to an ODE of the following form goes to $0$ as $trightarrowinfty$: $$y'+a(t)y=f(t)$$
with $a(t)$ and $f(t)$ continuous, $a(t) geq c>0$ and $limlimits_{trightarrowinfty}f(t)=0$.

We can find an integrating factor for this ODE, namely $mu(t)=e^{int a(t)}$, so that we find that any solution looks like $$y = e^{-int a(t)}int f(t)e^{int a(t)}$$ and from the property that $a(t) geq c>0$ we can conclude that $$|y| leq e^{-ct}int f(t)e^{int a(t)}$$
however I don't know how to continue from here. Any help would be appreciated.

Let $A(t) = int_0^t a(s), ds$. By method of integrating factors,
$$y(t) = e^{-A(t)}y(0) + e^{-A(t)}int_0^t e^{A(s)}f(s), ds$$ For $t ge 0$, the exponential $e^{-A(t)}$ is bounded by $e^{-ct}$, so the term $e^{-At}y(0)to 0$ as $t to infty$. Now $$e^{-A(t)}int_0^t e^{A(s)}f(s), ds = int_0^t f(s)e^{-int_s^t a(u), du}, ds$$ is bounded by $int_0^t lvert f(s)rvert e^{-c(t-s)}, ds$, which we'll show tends to $0$ as well. Fix a positive number $varepsilon$. Using the condition $limlimits_{tto infty} f(t) = 0$, we may choose a positive number $M$ such that for all $t ge M$, $lvert f(s)rvert < varepsilon$. For $t > M$, $$int_0^t lvert f(s)rvert e^{-c(t-s)}, ds = int_0^M lvert f(s)rvert e^{-c(t-s)}, ds + int_M^t lvert f(s)rvert e^{-c(t-s)}, ds le Ce^{-ct} + frac{varepsilon}{c}left[1-e^{-c(t-M)}right]$$ where $C$ is the constant $int_0^M lvert f(s)rvert e^{cs}, ds$. Taking the limit superior as $t to infty$ results in $$limsup_{tto infty} int_0^t lvert f(s)rvert e^{-c(t-s)}, ds le frac{varepsilon}{c}$$ As $varepsilon$ was arbitrary, $int_0^t lvert f(s)rvert e^{-c(t-s)}, ds$ tends to $0$ as $tto infty$, as desired.

Let $0 < varepsilon le 2$, let $T > 0$ be such that $|f(t)| le varepsilon$ for $t ge T$, and set $z(t) = |y(t)|^2$. Then for $t ge T$
$$
frac{d}{dt} z(t) = - 2a(t) z(t) + |f(t)y(t)| le -2cz(t) + varepsilon sqrt{z(t)}
le - 2 c z(t) +varepsilon (frac{1}{2c} + frac{c}{2} z(t))
$$
and therefore
$$
frac{d}{dt} z(t) le - cz(t) + frac{varepsilon}{2c} ,
$$
Now set $w(t) = e^{ct}z(t)$, then $frac{d}{dt} w(t) le frac{varepsilon e^{ct}}{2c}$.
Integrating this inequality from $T$ to $t > T$, it follows that
$$
w(t) le w(T) + frac{varepsilon}{2c^2}(e^{c(t-T)}-1)
$$
and therefore
$$
z(t) le e^{-ct}cdot const. + frac{varepsilon}{2c^2} ,
$$
Consequently $limsup_{t to infty} z(t) le frac{varepsilon}{2c^2}$.

Since $varepsilon$ was arbitrary, it follows that $y(t) to 0$.

This argument also works for vector valued differential equations, even nonlinear ones. An integrating factor for the equation is not needed.