Solve y(t)+2int_0^t e^{-2(t-tau)}y(tau)d tau=e^{-2t}

Given equation is :
$y(t)+2{\int }_0^t{e}^{-2(t-\tau )}y(\tau )d\tau =e^{-2t}$
To solve this equation use Laplace transform. Apply Laplace transform on both sides:
$L\{y(t)\}+2L\{{\int }_0^t{e}^{-2(t-\tau )}y(\tau )d\tau \}=L\left\{e^{-2t}\right\}$
The standard Laplace transform gives that $L\{{\int }_0^t{e}^{-2(t-\tau )}y(\tau )d\tau \}=L\left\{e^{-2t}\right\}\cdot L\{y(t)\}$
Let the Laplace transform of y(t) be Y(s)
Then the equation turns into:
$Y(s)+2L\left\{e^{-2t}\right\}\cdot Y(s)=L\left\{e^{-2t}\right\}$
Use the standard Laplace transform of $e^{-at}$, which is $\frac{1}{s+a}$
$\Rightarrow Y(s)+\frac{2}{s+2}Y(s)=\frac{1}{(s+2)}$
$\Rightarrow \left[\frac{(s+2+2)}{(s+2)}\right]Y(s)=\frac{1}{(s+2)}$
$\Rightarrow Y(s)=\frac{1}{(s+4)}$
Step 2
Now, apply the inverse Laplace transform on the obtained equation above.
$L^{-1}\{Y(s)\}=L^{-1}\left\{\frac{1}{(s+4)}\right\}$
Since the Laplace transform of y(t) is Y(s), inverse Laplace of Y(s) is y(t)
Similarly, when $L\left\{e^{(-at)}\right\}=\frac{1}{(s+a)}$, the inverse Laplace is: $L^{-1}\left\{\frac{1}{(s+a)}\right\}=e^{-at}$
Then, the resultant equation is:
$y(t)=e^{-4t}$
Therefore, the solution for the given equation is