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

In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $\frac{dy}{dt}$ also appears. Consider the following initial value problem, defined for t > 0:
$\frac{dy}{dt}+4{\int }_{0}^{t}y\left(t-w\right){e}^{-4w}dw=3,y\left(0\right)=0$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$Y\left(s\right)=L\left\{y\left(t\right)\right)\right\}-?$
b) Obtain the solution y(t).
y(t) - ?
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Step 1
Given,
a) Taking Laplace transform of the above equation
$L\left({y}^{\prime }\right)+4L\left({\int }_{0}^{t}y\left(t-w\right){e}^{-4w}dw\right)=L\left(3\right)$
$sy\left(s\right)-y\left(0\right)+4\frac{y\left(s\right)}{s+4}=\frac{3}{s}$
$y\left(s\right)\left[s+\frac{4}{s+4}\right]=\frac{3}{s}$
$y\left(s\right)\left[\frac{{s}^{2}+4s+4}{s+4}\right]=\frac{3}{s}$
$y\left(s\right)=\frac{3}{s}\left[\frac{s+4}{{s}^{2}+4s+4}\right]$
$=\frac{3}{s}\left[\frac{s+4}{{\left(s+2\right)}^{2}}\right]$
$y\left(s\right)=\frac{3\left(s+4\right)}{s{\left(s+2\right)}^{2}}$
Step 2
Now take, $\frac{3\left(s+4\right)}{s{\left(s+2\right)}^{2}}$ resolve into partial fractions
$\frac{3\left(s+4\right)}{s{\left(s+2\right)}^{2}}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{{\left(s+2\right)}^{2}}$
$\frac{3\left(s+4\right)}{s{\left(s+2\right)}^{2}}=\frac{A{\left(s+2\right)}^{2}+Bs\left(s+2\right)+Cs}{s{\left(s+2\right)}^{2}}$
$3\left(s+4\right)=A{\left(s+2\right)}^{2}+Bs\left(s+2\right)+Cs$
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
$\frac{3\left(s+4\right)}{s{\left(s+2\right)}^{2}}=\frac{3}{s}+\frac{-3}{s+2}+\frac{-3}{{\left(s+2\right)}^{2}}$