Let \(\displaystyle{y}={f{{\left({t},{y}\right)}}},{y}{\left({t}_{{{0}}}\right)}={y}_{{{0}}}\) be given IvP

The by Euler's method

\(\displaystyle{y}{\left({t}_{{{n}+{1}}}\right)}={y}_{{{n}+{1}}}={y}_{{{n}}}+{n}{f{{\left({t}_{{{n}}},{y}_{{{n}}}\right)}}}\), h is me step sire.

Step 2

Given \(\displaystyle{y}={7}{t}{e}^{{-{y}}}={f{{\left({t},{y}\right)}}}\) (say)

a) Then by Euler's method we have,

\(\displaystyle{y}_{{{n}+{1}}}={y}_{{{n}}}+{n}{f{{\left({t}_{{{n}}},{y}_{{{n}}}\right)}}}\), have \(\displaystyle{h}={0.2}\)

\(\displaystyle{t}_{{{0}}}={0},{y}_{{{0}}}={2}\)

\(\displaystyle\therefore{y}_{{{1}}}={y}_{{{0}}}+{0.2}{f{{\left({t}_{{{0}}},{y}_{{{0}}}\right)}}}={2}+{0.2}\cdot={2}={y}{\left({0.2}\right)}\)

\(\displaystyle\therefore{y}_{{{2}}}={y}_{{{1}}}+{0.2}{f{{\left({t}_{{{1}}},{y}_{{{1}}}\right)}}}={2}+{0.2}={y}{\left({0.189}\right)}={2.0378}={y}{\left({0.4}\right)}\)

\(\displaystyle\therefore{y}_{{{3}}}={y}_{{{2}}}+{0.2}{f{{\left({t}_{{{2}}},{y}_{{{2}}}\right)}}}={2.0378}+{0.2}{\left({0.3649}\right)}={2.11078}={y}{\left({0.6}\right)}\)

\(\displaystyle\therefore{y}_{{{4}}}={y}_{{{3}}}+{0.2}{f{{\left({t}_{{{3}}},{y}_{{{3}}}\right)}}}={2.11078}+{0.2}{\left({0.5088}\right)}={2.21254}={y}{\left({0.8}\right)}\)

\(\displaystyle\therefore{y}_{{{5}}}={y}_{{{4}}}+{0.2}{f{{\left({t}_{{{4}}},{y}_{{{4}}}\right)}}}={2.21254}+{0.2}{\left({0.6128}\right)}={2.3351}={y}{\left({1.0}\right)}\)

So we get:

\[\begin{array}{cc} k & t_{k} & y_{k} \\ 0 & 0 & 2 \\ 1 & 0.2 & 2 \\ 1 & 0.4 & 2.0378 \\ 3 & 0.6 & 2.11078 \\ 4 & 0.8 & 2.21251 \\ 5 & 1.0 & 2.3351\end{array}\]