Step 1

Given, \(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}\)

a) Taking Laplace transform of the above equation

\({L}{\left({y}'\right)}+{4}{L}{\left({\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}\right)}={L}{\left({3}\right)}\)

\({s}{y}{\left({s}\right)}-{y}{\left({0}\right)}+{4}\frac{{{y}{\left({s}\right)}}}{{{s}+{4}}}=\frac{3}{{s}}\)

\({y}{\left({s}\right)}{\left[{s}+\frac{4}{{{s}+{4}}}\right]}=\frac{3}{{s}}\)

\({y}{\left({s}\right)}{\left[\frac{{{s}^{2}+{4}{s}+{4}}}{{{s}+{4}}}\right]}=\frac{3}{{s}}\)

\({y}{\left({s}\right)}=\frac{3}{{s}}{\left[\frac{{{s}+{4}}}{{{s}^{2}+{4}{s}+{4}}}\right]}\)

\(=\frac{3}{{s}}{\left[\frac{{{s}+{4}}}{{{\left({s}+{2}\right)}^{2}}}\right]}\)

\({y}{\left({s}\right)}=\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}\)

Step 2

Now take, \(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}\) resolve into partial fractions

\(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{A}{{s}}+\frac{B}{{{s}+{2}}}+\frac{C}{{\left({s}+{2}\right)}^{2}}\)

\(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{{{A}{\left({s}+{2}\right)}^{2}+{B}{s}{\left({s}+{2}\right)}+{C}{s}}}{{{s}{\left({s}+{2}\right)}^{2}}}\)

\({3}{\left({s}+{4}\right)}={A}{\left({s}+{2}\right)}^{2}+{B}{s}{\left({s}+{2}\right)}+{C}{s}\)

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

\(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\)

\({y}{\left({s}\right)}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\)

i.e. , \({y}{\left({s}\right)}={L}{\left({y}{\left({t}\right)}\right)}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\)

Step 3

b) Take inverse Laplace transform,

\({L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\right)}\)

\(={L}^{ -{{1}}}{\left(\frac{3}{{s}}\right)}-{L}^{ -{{1}}}\frac{{{3}}}{{{s}+{2}}}-{L}^{ -{{1}}}\frac{{{3}}}{{\left({s}+{2}\right)}^{2}}\)

\(={3}{L}^{ -{{1}}}{\left(\frac{3}{{s}}\right)}-{3}{L}^{ -{{1}}}\frac{{{1}}}{{{s}+{2}}}-{3}{L}^{ -{{1}}}\frac{{{1}}}{{\left({s}+{2}\right)}^{2}}\)

\({L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={3}-{3}{e}^{{-{2}{t}}}-{3}{t}{e}^{{-{2}{t}}}\)

Therefore, \({y}{\left({t}\right)}={L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={3}-{3}{e}^{{-{2}{t}}}-{3}{t}{e}^{{-{2}{t}}}\)

Given, \(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}\)

a) Taking Laplace transform of the above equation

\({L}{\left({y}'\right)}+{4}{L}{\left({\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}\right)}={L}{\left({3}\right)}\)

\({s}{y}{\left({s}\right)}-{y}{\left({0}\right)}+{4}\frac{{{y}{\left({s}\right)}}}{{{s}+{4}}}=\frac{3}{{s}}\)

\({y}{\left({s}\right)}{\left[{s}+\frac{4}{{{s}+{4}}}\right]}=\frac{3}{{s}}\)

\({y}{\left({s}\right)}{\left[\frac{{{s}^{2}+{4}{s}+{4}}}{{{s}+{4}}}\right]}=\frac{3}{{s}}\)

\({y}{\left({s}\right)}=\frac{3}{{s}}{\left[\frac{{{s}+{4}}}{{{s}^{2}+{4}{s}+{4}}}\right]}\)

\(=\frac{3}{{s}}{\left[\frac{{{s}+{4}}}{{{\left({s}+{2}\right)}^{2}}}\right]}\)

\({y}{\left({s}\right)}=\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}\)

Step 2

Now take, \(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}\) resolve into partial fractions

\(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{A}{{s}}+\frac{B}{{{s}+{2}}}+\frac{C}{{\left({s}+{2}\right)}^{2}}\)

\(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{{{A}{\left({s}+{2}\right)}^{2}+{B}{s}{\left({s}+{2}\right)}+{C}{s}}}{{{s}{\left({s}+{2}\right)}^{2}}}\)

\({3}{\left({s}+{4}\right)}={A}{\left({s}+{2}\right)}^{2}+{B}{s}{\left({s}+{2}\right)}+{C}{s}\)

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

\(\frac{{{3}{\left({s}+{4}\right)}}}{{{s}{\left({s}+{2}\right)}^{2}}}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\)

\({y}{\left({s}\right)}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\)

i.e. , \({y}{\left({s}\right)}={L}{\left({y}{\left({t}\right)}\right)}=\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\)

Step 3

b) Take inverse Laplace transform,

\({L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={L}^{ -{{1}}}{\left(\frac{3}{{s}}+\frac{{-{3}}}{{{s}+{2}}}+\frac{{-{3}}}{{\left({s}+{2}\right)}^{2}}\right)}\)

\(={L}^{ -{{1}}}{\left(\frac{3}{{s}}\right)}-{L}^{ -{{1}}}\frac{{{3}}}{{{s}+{2}}}-{L}^{ -{{1}}}\frac{{{3}}}{{\left({s}+{2}\right)}^{2}}\)

\(={3}{L}^{ -{{1}}}{\left(\frac{3}{{s}}\right)}-{3}{L}^{ -{{1}}}\frac{{{1}}}{{{s}+{2}}}-{3}{L}^{ -{{1}}}\frac{{{1}}}{{\left({s}+{2}\right)}^{2}}\)

\({L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={3}-{3}{e}^{{-{2}{t}}}-{3}{t}{e}^{{-{2}{t}}}\)

Therefore, \({y}{\left({t}\right)}={L}^{ -{{1}}}{\left({y}{\left({s}\right)}\right)}={3}-{3}{e}^{{-{2}{t}}}-{3}{t}{e}^{{-{2}{t}}}\)