Let the function $g\colon\mathbb{R}\to\mathbb{R}$ be continuous for $t\ge t_0$, and consider the initial value problem
$$\begin{cases}y'+ay=g(t), & \\ y(t_0)=y_0. & \end{cases}$$
$(a)$ Verify that the solution for $t \ge t_0$ is given by $y=y_0e^{-at}+e^{-at} \int ^t_{t_0} e^{au}g(u)du$.
$(b)$ If $a>0$ and $\lim_{t\to \infty} g(t)=L$, show that $y$ tends $L/a$ as $t \to \infty$
Solution :
$(a)$ The differential equation $y'+ay=g(t); y(t_0)=y_0$ is of the first order and linear, hence solution is obtained as follows: $$\mathrm{e}^{at}\cdot y'+ \mathrm{e}^{at}\cdot ay = \mathrm{e}^{at}\cdot g(t),$$ $$\left(y\cdot \mathrm{e}^{at}\right)' = \mathrm{e}^{at}\cdot g(t),$$ $$y\cdot \mathrm{e}^{at}+c=\int \mathrm{e}^{at}g(t)dt,$$ $$y=\mathrm{e}^{-at}\left[\int \mathrm{e}^{at}g(t)dt+c\right].$$
Using initial condition $y(t_0)=y_0$
We get the result $y=y_0e^{-at}+e^{-at} \int ^t_{t_0} e^{au}g(u)du$.
(b) $y=\frac{y_0+\int ^t_{t_0} e^{au}g(u)du}{e^{at} }$
Apply $t\to \infty$ this is $\frac{\infty}{\infty}$ from So we use L-Hospital Rule
Therefor $\lim _{n \to \infty}y=\lim_{n \to \infty}\frac{e^{at}g(t)}{ae^{at}}=\frac{L}{a}$ as $\lim_{t\to \infty} g(t)=L$
Is my answer is correct?