Find the Laplace $\mathcal{L}(y(t))$ if $y'(t)-y(t)+\int_0^t y(u)e^{\alpha(t-u)}du=\alpha e^{\alpha t}$ How to find in terms of $y(0)$ and $\alpha$, the  Laplace transform $\mathcal{L}\{y(t)\}(s)$ if $$y'(t)-y(t)+\int_0^t y(u)e^{\alpha(t-u)}du=\alpha e^{\alpha t}$$
Comment: By the properties of Laplace transform, applying directly with $Y(s)=\mathcal{L}\{(y(t))\}(s) $
$$sY(s)-y(0)-Y(s)+Y(s)\mathcal{L}(e^{\alpha t})=\alpha\mathcal{L}(e^{\alpha t}) $$ here I'm using the convolution property, is it well applied? I mean is $$ \mathcal{L}\left(\int_0^t y(u)e^{\alpha(t-u)}du \right)=Y(s)\mathcal{L}(e^{\alpha t}) ?$$
 A: Note that if $f$ and $g$ be continous function on $\mathbb{R}^{+}$, so we can define the convolution between the functions $f$ and $g$ as $$f(t)\ast g(t)=\int_{0}^{t}f(u)g(t-u)\operatorname{du}.$$
Let's to make $f(u):=y(u)$ and let $g(t):=e^{at}$, so we have that $g(t-u)=e^{a(t-u)}$. Therefore, $$\color{blue}{y(u)\ast e^{at}}=\int_{0}^{t}y(u)e^{a(t-u)}\operatorname{du}.$$
On the other hand, by convolution's theorem we know that: if $f$ and $g$ be piecewise continous function on $\mathbb{R}^{+}$ and $f$ and $g$  be exponential order, so $$\mathcal{L}\{f\ast g\}=\mathcal{L}\{f(t)\}\cdot \mathcal{L}\{g(t)\}=F(s)G(s)$$
Then you can see that $$\color{red}{\mathcal{L}\left\{ \int_{0}^{t}y(u)e^{a(t-u)}\operatorname{du} \right\}}=\mathcal{L}\{y(u)\ast e^{at}\}\color{red}{=Y(s)\mathcal{L}\{e^{at}\}}.$$
Reviewing your comment, you may need to know what $$\color{blue}{e^{at}=\mathcal{L}^{-1}\left\{ \frac{1}{s-a}\right\},  \quad s\not=a}.$$
Finally, solve algebraically $Y(s)$ from the algebraic equation and apply $$\color{blue}{f(t)=\mathcal{L}^{-1}\{F(s)\}}.$$
