Solving integral equation $y(t) = e^t ( 1 + \int_0^t e^{-\tau} y(\tau)d \tau )$ with Laplace transform

Question

Solve the integral equation $$y(t) = e^t \bigg(1+ \int_0^t e^{-\tau}\ y(\tau) d \tau \bigg)$$

The apporach that stands out to me is using the convolution theorem and laplace transforms. Im able to arrive at the expression $$y(t) = e^t + e^t*y(t)$$ But Im unsure what do next. From my understanding you can't factor out $e^t$ in this case?

Answer 1

$$y(t) = e^t \bigg(1+ \int_0^t e^{-\tau}\ y(\tau) d \tau \bigg)$$ $$y'(t) = e^t \bigg(1+ \int_0^t e^{-\tau}\ y(\tau) d \tau \bigg) + y(t) = 2y(t) \implies y(t)=Ce^{2t}$$

Substitute into the original equation to find $C=1$.

$$Ce^{2t} = e^t \bigg(1+ C\int_0^t e^{\tau} d \tau \bigg)$$ $$Ce^{t} = 1+ C\left(e^t-1\right) \implies C=1$$

Answer 2

Let the Laplace of y(t) be Y(s).

Applying Laplace on both the sides we end up with

\begin{align} > Y(s)&= \frac{1}{s-1}+(L[e^t])(L[y(t)])\\ \end{align} \begin{align} > Y(s)&= \frac{1}{s-1}+\frac{1}{s-1}Y(s)\\ \end{align} Solving we end up in,

\begin{align} > Y(s)&= \frac{1}{s-2}\\ \end{align} Taking inverse Laplace we get, y(t)=e^2t