In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative frac{dy}{dt} also appears. Consid

Step 1
Given, $\frac{ dy }{ dt }+4{\int }_0^{t}y(t-w)e^{-4w}dw=3,y\left(0\right)=0$
a) Taking Laplace transform of the above equation
$L\left(y^{\prime }\right)+4L\left({\int }_0^{t}y(t-w)e^{-4w}dw\right)=L\left(3\right)$
$sy\left(s\right)-y\left(0\right)+4\frac{y\left(s\right)}{s+4}=\frac{3}{s}$
$y\left(s\right)[s+\frac{4}{s+4}]=\frac{3}{s}$
$y\left(s\right)\left[\frac{s^2+4s+4}{s+4}\right]=\frac{3}{s}$
$y\left(s\right)=\frac{3}{s}\left[\frac{s+4}{s^2+4s+4}\right]$
$=\frac{3}{s}\left[\frac{s+4}{{(s+2)}^2}\right]$
$y\left(s\right)=\frac{3(s+4)}{s{(s+2)}^2}$
Step 2
Now take, $\frac{3(s+4)}{s{(s+2)}^2}$ resolve into partial fractions
$\frac{3(s+4)}{s{(s+2)}^2}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{{(s+2)}^2}$
$\frac{3(s+4)}{s{(s+2)}^2}=\frac{A{(s+2)}^2+Bs(s+2)+Cs}{s{(s+2)}^2}$
$3(s+4)=A{(s+2)}^2+Bs(s+2)+Cs$
Put $s=0$
$12=4A+0+0$
A=3
Put $s=-2$
$6=0+0-2C$
$C=-3$
Consider coefficients of 's2'
$0=A+B$
$A=-B$ or $B=-A$
$B=-3$
Put the values in the equation
$\frac{3(s+4)}{s{(s+2)}^2}=\frac{3}{s}+\frac{-3}{s+2}+\frac{-3}{{(s+2)}^2}$