Given Initial Value Problem is:

\(\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}={3}-{2}{t}-{0},{5}{y},{y}{\left({0}\right)}={1},{t}_{{0}}={0}\)

and step size \(h = 0,05\)

Here \(\displaystyle f{{\left({t},{y}\right)}}={3}-{2}{t}-{0},{5}{y}\)

\(\displaystyle f{{\left({t}_{{{n}-{1}}},{y}_{{{n}-{1}}}\right)}}={3}-{2}{t}_{{{n}-{1}}}-{0},{5}{y}_{{{n}-{1}}}\)

\(\displaystyle={3}-{2}{\left\lbrace{t}_{{0}}+{\left({n}-{1}\right)}{h}\right\rbrace}-{0},{5}{y}_{{{n}-{1}}}\)

Now we have \(\displaystyle{t}_{{0}}={0}{\quad\text{and}\quad}{y}_{{0}}={1}\)

for \(n = 1\) \(\displaystyle\therefore f{{\left({t}_{{0}},{y}_{{0}}\right)}}={3}-{2}{\left\lbrace{0}+{\left({1}-{1}\right)}{h}\right\rbrace}-{0},{5}\)

\(\displaystyle={3}-{2}\times{0}-{0},{5}\times{1}\)

\(= 3 - 0,5\)

\(= 2,5\)

\(\displaystyle\therefore{y}_{{1}}={y}_{{0}}+{h} f{{\left({t}_{{0}},{y}_{{0}}\right)}}\)

\(\displaystyle={1}+{0},{05}\times{2},{5}\)

\(= 1 + 0,125\)

\(= 1,125\)

Now, \(\displaystyle{t}_{{1}}={t}_{{0}}+{\left({n}-{1}\right)}{h}={0}+{\left({2}-{1}\right)}\times{0},{05}={0},{05},{{f}_{{{0}{|}><}}{n}}={2}{{f}_{{{0}{|}><}}}\)

\(n = 2,\)

\(\displaystyle f{{\left({t}_{{1}},{y}_{{1}}\right)}}={3}-{2}{\left\lbrace{t}_{{0}}+{\left({n}-{1}\right)}{h}\right\rbrace}-{0},{5}{y}_{{1}}\)

\(\displaystyle={3}-{2}{\left\lbrace{0}+{\left({2}-{1}\right)}{0},{05}\right\rbrace}-{0},{5}\times{1},{125}\)

\(\displaystyle={3}-{2}\times{0},{05}-{0},{5625}\)

\(\displaystyle={3}-{0},{1}-{0},{05625}\)

\(= 2,3375\)

\(\displaystyle\therefore{y}_{{2}}={y}_{{1}}={h} f{{\left({t}_{{1}},{y}_{{1}}\right)}}\)

\(\displaystyle={1},{125}+{0},{05}\times{2},{3375}\)

\(\displaystyle={1},{125}+{0},{1169}{\left[{f}{o}{u}{r}\ {d}{e}{c}{i}{m}{a}{l}\ {p}{l}{a}{c}{e}{s}\right]}\)

\(= 1,2419\)

Again, \(\displaystyle{t}_{{2}}={t}_{{0}}+{\left({3}-{1}\right)}{h}\)

\(\displaystyle={0}+{2}\times{0},{05},{n}={3}\)

\(= 0,1\)

\(\displaystyle\therefore f{{\left({t}_{{2}},{y}_{{2}}\right)}}={3}-{2}{t}_{{2}}-{0},{5}{y}_{{2}}\)

\(\displaystyle={3}-{2}\times{0},{1}-{0},{5}\times{1},{2419}\)

\(\displaystyle={3}-{0},{2}-{0},{6210}{\left[{f}{o}{u}{r}\ {d}{e}{c}{i}{m}{a}{l}\ {s}{p}{l}{a}{c}{e}{s}\right]}\)

\(= 2,1790\)

Hence

\(\displaystyle{y}_{{3}}={y}_{{2}}+{h} f{{\left({t}_{{2}},{y}_{{2}}\right)}}\)

\(\displaystyle={1},{2419}+{0},{05}\times{2},{1790}\)

\(= 1,2419\ +\ 0,1090\)

\(= 1,3509\)

Here \(t_{2} = 0,1\ \text{and at}\ t_{2} = 0.1,\ y_{3} = 1.3509\)

\(\displaystyle\therefore{y}_{{3}}={y}{\left({0.1}\right)}={1.3509}\)